Search: keyword:new
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A371970
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Exponents k such that the binary expansion of 3^k has an even number of ones.
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0
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1, 2, 3, 5, 6, 8, 9, 12, 14, 17, 18, 21, 23, 24, 25, 26, 27, 31, 32, 33, 35, 37, 38, 39, 40, 42, 44, 45, 47, 51, 52, 55, 57, 58, 59, 60, 61, 64, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 81, 96, 99, 102, 104, 105, 106, 109, 112, 116, 127, 131, 132, 133, 134, 135, 136
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OFFSET
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1,2
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LINKS
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MAPLE
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q:= n-> is(add(i, i=Bits[Split](3^n))::even):
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MATHEMATICA
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PROG
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(PARI) is_a371970(k) = hammingweight(3^k)%2 == 0
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CROSSREFS
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KEYWORD
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nonn,base,easy,new
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AUTHOR
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STATUS
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approved
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A372237
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a(0) = 4; to obtain a(k), write out the base-(2^k) expansion of a(k-1), bump to base 2^(k+1), then subtract 1.
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0
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4, 15, 26, 49, 96, 191, 318, 573, 1084, 2107, 4154, 8249, 16440, 32823, 65590, 131125, 262196, 524339, 1048626, 2097201, 4194352, 8388655, 16777262, 33554477, 67108908, 134217771, 268435498, 536870953, 1073741864, 2147483687, 4294967334, 8589934629, 17179869220
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OFFSET
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0,1
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COMMENTS
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Applying to the proof of the usual Goodstein's theorem to the ordinal number omega^omega shows that: for no matter what initial value and no matter what increasing sequence of bases b(0), b(1), ... with b(0) >= 2, the (weak) Goodstein sequence eventually terminates with 0. Here b(k) = 2^(k+1).
Sequence terminates at a(2^(2^70+70) + 2^70 + 68) = 0.
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LINKS
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FORMULA
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a(k) = 2^(k+2) + 68 - k for 5 <= k <= 68. The base-(2^(k+1)) expansion of a(k) consists of two digits 2 and 68 - k.
a(k) = 2^(k+1) + 2^70 + 68 - k for 69 <= 2^70 + 68. The base-(2^(k+1)) expansion of a(k) consists of two digits 1 and 2^70 + 68 - k.
a(k) = 2^(2^70+70) + 2^70 + 68 - k for 2^70 + 69 <= k <= 2^(2^70+70) + 2^70 + 68. The base-(2^(k+1)) expansion of a(k) consists of a single digit 2^(2^70+70) + 2^70 + 68 - k.
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EXAMPLE
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a(0) = 100_2 = 4;
a(1) = 100_4 - 1 = 15 = 33_4;
a(2) = 33_8 - 1 = 26 = 32_8;
a(3) = 32_16 - 1 = 49 = 31_16;
a(4) = 31_32 - 1 = 96 = 30_32;
a(5) = 30_64 - 1 = 191 = (2,63)_64.
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PROG
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(PARI) A372237_first_N_terms(N) = my(v=vector(N+1)); v[1] = 4; for(i=1, N, v[i+1] = fromdigits(digits(v[i], 2^i), 2^(i+1))-1); v
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CROSSREFS
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KEYWORD
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nonn,easy,fini,new
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AUTHOR
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STATUS
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approved
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A372261
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Number T(n,k,j) of acyclic orientations of the complete tripartite graph K_{n,k,j}; triangle of triangles T(n,k,j), n>=0, k=0..n, j=0..k, read by rows.
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1, 1, 2, 6, 1, 4, 18, 14, 78, 426, 1, 8, 54, 46, 330, 2286, 230, 1902, 15402, 122190, 1, 16, 162, 146, 1374, 12090, 1066, 10554, 101502, 951546, 6902, 76110, 822954, 8724078, 90768378, 1, 32, 486, 454, 5658, 63198, 4718, 57054, 657210, 7290942, 41506, 525642, 6495534, 78463434, 928340190
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OFFSET
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0,3
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COMMENTS
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An acyclic orientation is an assignment of a direction to each edge such that no cycle in the graph is consistently oriented. Stanley showed that the number of acyclic orientations of a graph G is equal to the absolute value of the chromatic polynomial X_G(q) evaluated at q=-1.
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LINKS
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EXAMPLE
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Triangle of triangles T(n,k,j) begins:
1;
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1;
2, 6;
;
1;
4, 18;
14, 78, 426;
;
1;
8, 54;
46, 330, 2286;
230, 1902, 15402, 122190;
;
...
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MAPLE
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g:= proc(n) option remember; `if`(n=0, 1, add(
expand(x*g(n-j))*binomial(n-1, j-1), j=1..n))
end:
T:= proc() option remember; local q, l, b; q, l, b:= -1, [args],
proc(n, j) option remember; `if`(j=1, mul(q-i, i=0..n-1)*
(q-n)^l[1], add(b(n+m, j-1)*coeff(g(l[j]), x, m), m=0..l[j]))
end; abs(b(0, nops(l)))
end:
seq(seq(seq(T(n, k, j), j=0..k), k=0..n), n=0..5);
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CROSSREFS
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T(n,k,0) for k=0..9 give: A000012, A000079, A027649, A027650, A027651, A283811, A283812, A283813, A284032, A284033.
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KEYWORD
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AUTHOR
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STATUS
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approved
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A372254
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Number A(n,k) of acyclic orientations of the complete tripartite graph K_{n,n,k}; square array A(n,k), n>=0, k>=0, read by antidiagonals.
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0
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1, 1, 2, 1, 6, 14, 1, 18, 78, 230, 1, 54, 426, 1902, 6902, 1, 162, 2286, 15402, 76110, 329462, 1, 486, 12090, 122190, 822954, 4553166, 22934774, 1, 1458, 63198, 951546, 8724078, 61796298, 381523758, 2193664790, 1, 4374, 327306, 7290942, 90768378, 823457454, 6241779786, 42700751022, 276054834902
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OFFSET
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0,3
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COMMENTS
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An acyclic orientation is an assignment of a direction to each edge such that no cycle in the graph is consistently oriented. Stanley showed that the number of acyclic orientations of a graph G is equal to the absolute value of the chromatic polynomial X_G(q) evaluated at q=-1.
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LINKS
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EXAMPLE
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Square array A(n,k) begins:
1, 1, 1, 1, 1, 1, ...
2, 6, 18, 54, 162, 486, ...
14, 78, 426, 2286, 12090, 63198, ...
230, 1902, 15402, 122190, 951546, 7290942, ...
6902, 76110, 822954, 8724078, 90768378, 928340190, ...
329462, 4553166, 61796298, 823457454, 10779805722, 138779942046, ...
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MAPLE
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g:= proc(n) option remember; `if`(n=0, 1, add(
expand(x*g(n-j))*binomial(n-1, j-1), j=1..n))
end:
A:= proc(n, k) option remember; local q, l, b; q, l, b:= -1, [n$2, k],
proc(n, j) option remember; `if`(j=1, mul(q-i, i=0..n-1)*
(q-n)^l[1], add(b(n+m, j-1)*coeff(g(l[j]), x, m), m=0..l[j]))
end; abs(b(0, 3))
end:
seq(seq(A(n, d-n), n=0..d), d=0..9);
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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A371097
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Array A read by upward antidiagonals in which the entry A(n,k) in row n and column k is defined by A(n, k) = A371092(A371095(n, k)), n,k >= 1.
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1, 1, 2, 1, 2, 3, 1, 3, 1, 4, 1, 3, 1, 5, 5, 1, 1, 1, 2, 4, 6, 1, 1, 1, 3, 5, 8, 7, 1, 1, 1, 3, 2, 12, 2, 8, 1, 1, 1, 1, 3, 18, 2, 11, 9, 1, 1, 1, 1, 3, 27, 3, 9, 7, 10, 1, 1, 1, 1, 1, 21, 3, 7, 2, 14, 11, 1, 1, 1, 1, 1, 16, 1, 2, 2, 21, 4, 12, 1, 1, 1, 1, 1, 23, 1, 2, 3, 8, 6, 17, 13, 1, 1, 1, 1, 1, 18, 1, 3, 3, 12, 9, 4, 10, 14
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OFFSET
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1,3
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COMMENTS
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LINKS
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FORMULA
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EXAMPLE
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Array begins:
n\k| 1 2 3 ...
---+--------------------------------------------------------------
1 | 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17,
2 | 1, 2, 1, 5, 4, 8, 2, 11, 7, 14, 4, 17, 10, 20, 1, 23, 13,
3 | 1, 3, 1, 2, 5, 12, 2, 9, 2, 21, 6, 4, 14, 30, 1, 18, 10,
4 | 1, 3, 1, 3, 2, 18, 3, 7, 2, 8, 9, 5, 21, 45, 1, 26, 14,
5 | 1, 1, 1, 3, 3, 27, 3, 2, 3, 12, 1, 2, 8, 17, 1, 39, 21,
6 | 1, 1, 1, 1, 3, 21, 1, 2, 3, 18, 1, 3, 12, 4, 1, 30, 8,
7 | 1, 1, 1, 1, 1, 16, 1, 3, 1, 27, 1, 3, 18, 5, 1, 44, 12,
8 | 1, 1, 1, 1, 1, 23, 1, 3, 1, 21, 1, 1, 27, 2, 1, 66, 18,
9 | 1, 1, 1, 1, 1, 18, 1, 1, 1, 16, 1, 1, 21, 3, 1, 99, 27,
10 | 1, 1, 1, 1, 1, 26, 1, 1, 1, 23, 1, 1, 16, 3, 1, 75, 21,
11 | 1, 1, 1, 1, 1, 39, 1, 1, 1, 18, 1, 1, 23, 1, 1, 28, 16,
12 | 1, 1, 1, 1, 1, 30, 1, 1, 1, 26, 1, 1, 18, 1, 1, 42, 23,
13 | 1, 1, 1, 1, 1, 44, 1, 1, 1, 39, 1, 1, 26, 1, 1, 63, 18,
14 | 1, 1, 1, 1, 1, 66, 1, 1, 1, 30, 1, 1, 39, 1, 1, 48, 26,
15 | 1, 1, 1, 1, 1, 99, 1, 1, 1, 44, 1, 1, 30, 1, 1, 71, 39,
16 | 1, 1, 1, 1, 1, 75, 1, 1, 1, 66, 1, 1, 44, 1, 1, 54, 30,
17 | 1, 1, 1, 1, 1, 28, 1, 1, 1, 99, 1, 1, 66, 1, 1, 80, 44,
18 | 1, 1, 1, 1, 1, 42, 1, 1, 1, 75, 1, 1, 99, 1, 1, 120, 66,
19 | 1, 1, 1, 1, 1, 63, 1, 1, 1, 28, 1, 1, 75, 1, 1, 180, 99,
20 | 1, 1, 1, 1, 1, 48, 1, 1, 1, 42, 1, 1, 28, 1, 1, 270, 75,
21 | 1, 1, 1, 1, 1, 71, 1, 1, 1, 63, 1, 1, 42, 1, 1, 405, 28,
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PROG
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(PARI)
up_to = 105;
R(n) = { n = 1+3*n; n>>valuation(n, 2); };
A371095sq(n, k) = if(1==n, 8*k-7, R(A371095sq(n-1, k)));
A371097sq(n, k) = A371092(A371095sq(n, k));
A371097list(up_to) = { my(v = vector(up_to), i=0); for(a=1, oo, for(col=1, a, i++; if(i > up_to, return(v)); v[i] = A371097sq((a-(col-1)), col))); (v); };
v371097 = A371097list(up_to);
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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A371095
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Array A read by upward antidiagonals in which the entry A(n,k) in row n and column k is defined by A(1, k) = 8*k-7, and A(n+1, k) = R(A(n, k)), n,k >= 1, where Reduced Collatz function R(n) gives the odd part of 3n+1.
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0
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1, 1, 9, 1, 7, 17, 1, 11, 13, 25, 1, 17, 5, 19, 33, 1, 13, 1, 29, 25, 41, 1, 5, 1, 11, 19, 31, 49, 1, 1, 1, 17, 29, 47, 37, 57, 1, 1, 1, 13, 11, 71, 7, 43, 65, 1, 1, 1, 5, 17, 107, 11, 65, 49, 73, 1, 1, 1, 1, 13, 161, 17, 49, 37, 55, 81, 1, 1, 1, 1, 5, 121, 13, 37, 7, 83, 61, 89, 1, 1, 1, 1, 1, 91, 5, 7, 11, 125, 23, 67, 97
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OFFSET
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1,3
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LINKS
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EXAMPLE
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Array begins:
n\k| 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
---+------------------------------------------------------------------------
1 | 1, 9, 17, 25, 33, 41, 49, 57, 65, 73, 81, 89, 97, 105, 113, 121,
2 | 1, 7, 13, 19, 25, 31, 37, 43, 49, 55, 61, 67, 73, 79, 85, 91,
3 | 1, 11, 5, 29, 19, 47, 7, 65, 37, 83, 23, 101, 55, 119, 1, 137,
4 | 1, 17, 1, 11, 29, 71, 11, 49, 7, 125, 35, 19, 83, 179, 1, 103,
5 | 1, 13, 1, 17, 11, 107, 17, 37, 11, 47, 53, 29, 125, 269, 1, 155,
6 | 1, 5, 1, 13, 17, 161, 13, 7, 17, 71, 5, 11, 47, 101, 1, 233,
7 | 1, 1, 1, 5, 13, 121, 5, 11, 13, 107, 1, 17, 71, 19, 1, 175,
8 | 1, 1, 1, 1, 5, 91, 1, 17, 5, 161, 1, 13, 107, 29, 1, 263,
9 | 1, 1, 1, 1, 1, 137, 1, 13, 1, 121, 1, 5, 161, 11, 1, 395,
10 | 1, 1, 1, 1, 1, 103, 1, 5, 1, 91, 1, 1, 121, 17, 1, 593,
11 | 1, 1, 1, 1, 1, 155, 1, 1, 1, 137, 1, 1, 91, 13, 1, 445,
12 | 1, 1, 1, 1, 1, 233, 1, 1, 1, 103, 1, 1, 137, 5, 1, 167,
13 | 1, 1, 1, 1, 1, 175, 1, 1, 1, 155, 1, 1, 103, 1, 1, 251,
14 | 1, 1, 1, 1, 1, 263, 1, 1, 1, 233, 1, 1, 155, 1, 1, 377,
15 | 1, 1, 1, 1, 1, 395, 1, 1, 1, 175, 1, 1, 233, 1, 1, 283,
16 | 1, 1, 1, 1, 1, 593, 1, 1, 1, 263, 1, 1, 175, 1, 1, 425,
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PROG
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(PARI)
up_to = 91;
R(n) = { n = 1+3*n; n>>valuation(n, 2); };
A371095sq(n, k) = if(1==n, 8*k-7, R(A371095sq(n-1, k)));
A371095list(up_to) = { my(v = vector(up_to), i=0); for(a=1, oo, for(col=1, a, i++; if(i > up_to, return(v)); v[i] = A371095sq((a-(col-1)), col))); (v); };
v371095 = A371095list(up_to);
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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A372056
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Smallest prime obtained by appending one or more 3's to n, or -1 if no such prime exists.
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+0
0
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13, 23, -1, 43, 53, -1, 73, 83, -1, 103, 113, -1, 1333333333333333, 1433, -1, 163, 173, -1, 193, 20333, -1, 223, 233, -1, 2533333333, 263, -1, 283, 293, -1, 313, 323333, -1, 3433, 353, -1, 373, 383, -1
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OFFSET
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1,1
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COMMENTS
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Next term is 40 followed by 483 3's and is too large to display here (see the b-file).
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LINKS
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EXAMPLE
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For n = 13, a(13) = 1333333333333333 is a prime (but 133,1333,13333 etc. are not primes).
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CROSSREFS
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KEYWORD
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base,sign,new
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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A371602
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Taxicab numbers that are sandwiched between squarefree numbers.
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+0
0
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4104, 32832, 39312, 110808, 171288, 262656, 314496, 373464, 513000, 886464, 1016496, 1075032, 1195112, 1331064, 1370304, 1407672, 1609272, 1728216, 1734264, 1774656, 2101248, 2515968, 2864288, 2987712, 2991816, 3511872, 3512808, 3551112, 4104000, 4342914, 4467528, 4511808, 4607064
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OFFSET
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1,1
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COMMENTS
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All terms are even numbers.
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LINKS
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Christian Boyer, Les nombres Taxicabs, in Dossier Pour La Science, pp. 26-28, Volume 59 (Jeux math') April/June 2008 Paris.
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EXAMPLE
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4104 = 2^3 * 3^3 * 19 (between 4103 = 11 * 373 and 4105 = 5 * 821).
32832 = 2^6 * 3^3 *19 (between 32831 and 32833, which are twin primes).
39312 = 2^4 * 3^3 * 7 * 13 (between 39311 = 19 * 2069 and 39313, which is prime).
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MATHEMATICA
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Select[Range[300000], And @@ SquareFreeQ /@ (# + {-1, 1}) && Length[PowersRepresentations[#, 2, 3]] > 1 &] (* Amiram Eldar, Mar 29 2024 *)
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CROSSREFS
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KEYWORD
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nonn,new
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AUTHOR
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STATUS
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approved
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A371611
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Number of 2*n-sided cycles with the property that one makes the same number of left and right turns while following its edges.
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+0
0
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4, 36, 1240, 73240, 7171176, 1016813448, 198480110880, 50752206180576, 16460660622560680, 6595414427636900536, 3198428240666246044704, 1845848150787599809368856, 1250049326783769438348496480, 981653074459964543314138858320
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OFFSET
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2,1
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COMMENTS
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Cycles that differ by rotation or reflection are counted separately. By "n-sided cycles" we mean the cycles that can be drawn by connecting n equally spaced points on a circle (possibly self-intersecting).
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LINKS
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FORMULA
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a(n) is always divisible by 2*n, because the considered polygons cannot have rotational symmetry.
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CROSSREFS
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KEYWORD
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nonn,new
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AUTHOR
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STATUS
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approved
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A371598
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a(n) = (Product_{i=1..n} Fibonacci(i)) mod Fibonacci(n + 1).
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+0
0
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0, 1, 2, 1, 6, 6, 12, 2, 15, 16, 0, 49, 299, 220, 882, 252, 2176, 166, 495, 5720, 5251, 6065, 28224, 41650, 106947, 113288, 256737, 173841, 26840, 25379, 444150, 347278, 1834953, 8709610, 4046544, 2653673, 31127545, 47532000, 50717205, 147239197, 97769672, 37543458
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OFFSET
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1,3
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LINKS
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FORMULA
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EXAMPLE
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a(2) = (1 * 1) mod 2 = 1.
a(3) = (1 * 1 * 2) mod 3 = 2.
a(4) = (1 * 1 * 2 * 3) mod 5 = 1.
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MATHEMATICA
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a[n_] := Mod[Fibonorial[n], Fibonacci[n + 1]]; Array[a, 50] (* Amiram Eldar, Mar 29 2024 *)
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PROG
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(Python)
from sympy import fibonacci
def a(n):
a_n = 1
mod = fibonacci(n + 1)
for i in range(1, n + 1):
a_n = (a_n * fibonacci(i)) % mod
return a_n
(PARI) a(n) = my(f=fibonacci(n+1)); lift(prod(k=1, n, Mod(fibonacci(k), f))); \\ Michel Marcus, Apr 03 2024
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CROSSREFS
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KEYWORD
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nonn,new
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AUTHOR
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STATUS
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approved
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