Search: keyword:new
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A371715
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G.f. A(x) = Sum_{n>=0} a(n)*x^n where a(n) = Sum_{k=0..n} ( [x^k] A(2*x)^(n+1) (mod 2^(n+1)) ) for n >= 0 with a(0) = 1.
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0
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1, 1, 7, 17, 55, 113, 135, 321, 2103, 3217, 8295, 18145, 33687, 74289, 128455, 321153, 870135, 1201105, 3696423, 6715937, 17466967, 31467889, 72111239, 173224385, 382697911, 552100625, 1442627047, 2558102881, 3487807767, 11651874993, 21616075591, 47612431617, 108598963319
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OFFSET
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0,3
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COMMENTS
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What is the limit a(n)/(n*2^n) ? Does it exist ?. - Vaclav Kotesovec, May 03 2024
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LINKS
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EXAMPLE
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G.f.: A(x) = 1 + x + 7*x^2 + 17*x^3 + 55*x^4 + 113*x^5 + 135*x^6 + 321*x^7 + 2103*x^8 + 3217*x^9 + 8295*x^10 + 18145*x^11 + 33687*x^12 + ...
The table of coefficients of x^k in A(2*x)^n begins:
n=1: [1, 2, 28, 136, 880, 3616, 8640, ...];
n=2: [1, 4, 60, 384, 3088, 18368, 99520, ...];
n=3: [1, 6, 96, 752, 6960, 50592, 350848, ...];
n=4: [1, 8, 136, 1248, 12848, 107520, 864000, ...];
n=5: [1, 10, 180, 1880, 21120, 197312, 1765760, ...];
n=6: [1, 12, 228, 2656, 32160, 329088, 3210624, ...];
n=7: [1, 14, 280, 3584, 46368, 512960, 5383168, ...];
n=8: [1, 16, 336, 4672, 64160, 760064, 8500480, ...];
...
from which each term in row n may be reduced modulo 2^n to form the following triangle (trailing zeros suppressed):
A(2*x) (mod 2): [1];
A(2*x)^2 (mod 2^2): [1, 0];
A(2*x)^3 (mod 2^3): [1, 6, 0];
A(2*x)^4 (mod 2^4): [1, 8, 8, 0];
A(2*x)^5 (mod 2^5): [1, 10, 20, 24, 0];
A(2*x)^6 (mod 2^6): [1, 12, 36, 32, 32, 0];
A(2*x)^7 (mod 2^7): [1, 14, 24, 0, 32, 64, 0];
A(2*x)^8 (mod 2^8): [1, 16, 80, 64, 160, 0, 0, 0];
A(2*x)^9 (mod 2^9): [1, 18, 396, 296, 464, 224, 320, 384, 0];
A(2*x)^10 (mod 2^10): [1, 20, 460, 192, 624, 832, 64, 512, 512, 0];
A(2*x)^11 (mod 2^11): [1, 22, 528, 784, 80, 1120, 1664, 1536, 1536, 1024, 0];
A(2*x)^12 (mod 2^12): [1, 24, 600, 2592, 3920, 3328, 2048, 2048, 3584, 0, 0, 0]; ...
The row sums of the above triangle form this sequence.
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PROG
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(PARI) /* Returns vector A of N terms */
{N = 40; A=vector(N); A[1]=1; for(n=2, #A, A[n]=1; A[n] = sum(k=0, n-1, ( polcoeff( Ser(A)^n, k)*2^k )%(2^n) ) ); A}
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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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A371709
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Expansion of g.f. A(x) satisfying A( x*A(x)^2 + x*A(x)^3 ) = A(x)^3.
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1, 1, 1, 2, 6, 16, 39, 99, 271, 764, 2157, 6128, 17658, 51534, 151635, 448962, 1337493, 4008040, 12072594, 36524898, 110943633, 338218626, 1034509917, 3173811240, 9763898994, 30113782641, 93094164244, 288415278638, 895332445053, 2784580242557, 8675408291598, 27072326322939
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OFFSET
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1,4
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COMMENTS
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a(3^n) == 1 (mod 3) for n >= 0.
a(2*3^n) == 1 (mod 3) for n >= 0.
a(n) == 2 (mod 3) iff n is the sum of 2 distinct powers of 3 (A038464).
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LINKS
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FORMULA
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G.f. A(x) = Sum_{n>=1} a(n)*x^n satisfies the following formulas.
(1) A( x*A(x)^2*(1 + A(x)) ) = A(x)^3.
(2) A(x) = x * Product_{n>=0} (1 + A(x)^(3^n)).
(3) A(x) = Series_Reversion( x / Product_{n>=0} (1 + x^(3^n)) ).
a(n) ~ c * d^n / n^(3/2), where d = 3.2753449994351908157330968510747739... and c = 0.1559869008021616116037651076359... - Vaclav Kotesovec, May 03 2024
The radius of convergence r of g.f. A(x) and A(r) satisfy A(r) = 1 / Sum_{n>=0} 3^n * A(r)^(3^n-1) / (1 + A(r)^(3^n)) and r = A(r) / Product_{n>=0} (1 + A(r)^(3^n)), where r = 0.30531134893345362211... = 1/d (d is given above) and A(r) = 0.600582105427215700175254768411726892599... - Paul D. Hanna, May 03 2024
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EXAMPLE
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G.f. A(x) = x + x^2 + x^3 + 2*x^4 + 6*x^5 + 16*x^6 + 39*x^7 + 99*x^8 + 271*x^9 + 764*x^10 + 2157*x^11 + 6128*x^12 + 17658*x^13 + 51534*x^14 + 151635*x^15 + 448962*x^16 + ...
where A( x*A(x)^2*(1 + A(x)) ) = A(x)^3.
RELATED SERIES.
A(x)^2 = x^2 + 2*x^3 + 3*x^4 + 6*x^5 + 17*x^6 + 48*x^7 + 126*x^8 + 332*x^9 + 918*x^10 + 2616*x^11 + 7504*x^12 + ...
A(x)^3 = x^3 + 3*x^4 + 6*x^5 + 13*x^6 + 36*x^7 + 105*x^8 + 292*x^9 + 801*x^10 + 2256*x^11 + 6515*x^12 + 18981*x^13 + ...
A(x)^2 + A(x)^3 = x^2 + 3*x^3 + 6*x^4 + 12*x^5 + 30*x^6 + 84*x^7 + 231*x^8 + 624*x^9 + 1719*x^10 + 4872*x^11 + 14019*x^12 + 40599*x^13 + ...
Let B(x) be the series reversion of g.f. A(x), B(A(x)) = x, then
B(x) * (1+x)/(1+x^3) = x - 2*x^4 + 3*x^7 - 5*x^10 + 7*x^13 - 9*x^16 + 12*x^19 - 15*x^22 + 18*x^25 - 23*x^28 + ... + (-1)^n*A005704(n)*x^(3*n+1) + ...
where A005704 is the number of partitions of 3*n into powers of 3.
We can show that g.f. A(x) = A( x*A(x)^2*(1 + A(x)) )^(1/3) satisfies
(2) A(x) = x * Product_{n>=0} (1 + A(x)^(3^n))
by substituting x*A(x)^2*(1 + A(x)) for x in (2) to obtain
A(x)^3 = x * A(x)^2*(1 + A(x)) * Product_{n>=1} (1 + A(x)^(3^n))
which is equivalent to formula (2).
SPECIFIC VALUES.
A(3/10) = 0.526165645044542830201162330432965674027415264612114520...
A(1/4) = 0.353259384374080248921564026412797625837830114153200664...
A(1/5) = 0.255218141344695821239609680309162895225297482063273545...
A(t) = 1/2 and A(t*3/8) = 1/8 at t = (1/2)/Product_{n>=0} (1 + 1/2^(3^n)) = 0.295718718466711580562679377308518930409875701753934072...
A(t) = 1/3 and A(t*4/27) = 1/27 at t = (1/3)/Product_{n>=0} (1 + 1/3^(3^n)) = 0.241059181496179959557718992589733756750585121455883861...
A(t) = 1/4 and A(t*5/64) = 1/64 at t = (1/4)/Product_{n>=0} (1 + 1/4^(3^n)) = 0.196922325724019432212969925740117827612003158137366017...
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PROG
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(PARI) /* Using series reversion of x/Product_{n>=0} (1 + x^(3^n)) */
{a(n) = my(A); A = serreverse( x/prod(k=0, ceil(log(n)/log(3)), (1 + x^(3^k) +x*O(x^n)) ) ); polcoeff(A, n)}
for(n=1, 35, print1(a(n), ", "))
(PARI) /* Using A(x)^3 = A( x*A(x)^2 + x*A(x)^3 ) */
{a(n) = my(A=[1], F); for(i=1, n, A = concat(A, 0); F = x*Ser(A);
A[#A] = polcoeff( subst(F, x, x*F^2 + x*F^3 ) - F^3, #A+2) ); A[n]}
for(n=1, 35, print1(a(n), ", "))
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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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A371505
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Number of sub-monoids of the monoid of uniform block permutations of size n that contain the symmetric group S_n.
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0
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1, 2, 3, 6, 10, 31, 63, 287, 1099, 8640, 62658, 1546891, 29789119, 2525655957
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OFFSET
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1,2
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COMMENTS
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Also equal to the number of anti-chains in the poset of integer partitions of k not equal to (1^k) where mu < lambda iff mu is coarser than lambda and r(mu) >= r(lambda) where r(lambda) = smallest part of lambda not equal to 1.
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LINKS
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EXAMPLE
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a(3) = 3 because the uniform block permutations of size 3; S_3; and the monoid consisting of S_3 and the element with one block are the only three sub-monoids.
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KEYWORD
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nonn,more,new
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AUTHOR
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STATUS
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approved
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A372304
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Binomial transform of the Gray code sequence.
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0, 1, 5, 14, 36, 92, 228, 536, 1200, 2608, 5624, 12224, 26920, 59824, 133024, 293504, 638816, 1367488, 2877728, 5962112, 12198000, 24748192, 50041312, 101366272, 206655440, 425423136, 885598720, 1863228544, 3953671808, 8436151552, 18042976640, 38567833600
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OFFSET
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0,3
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COMMENTS
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Also second binomial transform of A109629.
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LINKS
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FORMULA
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a(n) = Sum_{k=0..n} binomial(n,k) * A003188(k).
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MAPLE
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g:= proc(n) g(n):= Bits[Xor](n, iquo(n, 2)) end:
a:= n-> add(binomial(n, k)*g(k), k=0..n):
seq(a(n), n=0..40);
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PROG
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(Python)
from math import comb
def A372304(n): return sum(comb(n, k)*(k^k>>1) for k in range(n+1)) # Chai Wah Wu, May 02 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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1, 2, 3, 4, 8, 17, 23, 29, 38, 39, 44, 56, 57, 58, 91, 114, 145, 147, 156, 168, 182, 208, 219, 239, 277, 297, 300, 307, 331, 360, 367, 442, 452, 477, 487, 492, 507, 513, 568, 571, 614, 893, 963, 1275, 1283, 1288, 1440, 1563, 1702, 1957, 2019, 2440, 2471, 2566, 3004
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OFFSET
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1,2
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COMMENTS
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a(n)/2 are the differences at supporting points of the upper envelope of the scatter band of the deviation of the binary weight of 3^k from half the length of the corresponding binary number.
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LINKS
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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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0, 1, 3, 5, 11, 27, 71, 119, 140, 158, 198, 218, 441, 537, 538, 868, 1092, 2128, 2294, 2343, 2811, 2911, 3849, 4003, 4655, 5079, 5279, 5920, 6269, 6603, 10181, 10574, 12801, 12803, 15563, 15784, 16054, 16253, 17127, 18257, 20187, 21934, 34633, 49209, 76791, 78938
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OFFSET
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1,3
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COMMENTS
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These are the k-values of the upper envelope of the scatter band of the deviation of the binary weight of 3^k from half the length of the corresponding binary number. The corresponding differences are given in A372100.
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LINKS
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PROG
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(PARI) a372099(upto) = {my(dm=oo); for (k=0, upto, my (p=3^k, h=hammingweight(p), b=#binary(p)/2, d=b-h); if (d<dm, print1(k, ", "); dm=d))};
a372099(80000)
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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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-1, 0, 1, 2, 4, 7, 8, 12, 15, 18, 25, 26, 30, 51, 75, 78, 84, 129, 133, 148, 170, 180, 183, 189, 209, 265, 279, 285, 287, 336, 369, 388, 406, 412, 445, 469, 496, 581, 711, 737, 741, 742, 873, 939, 994, 1044, 1078, 1111, 1157, 1158, 1492, 1636, 1767, 1914, 1933
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OFFSET
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1,4
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COMMENTS
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a(n)/2 are the negated differences at supporting points of the lower envelope of the scatter band of the deviation of the binary weight of 3^k from half the length of the corresponding binary number.
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LINKS
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CROSSREFS
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KEYWORD
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sign,new
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AUTHOR
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STATUS
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approved
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0, 2, 4, 7, 16, 24, 40, 49, 53, 102, 104, 126, 174, 226, 379, 768, 831, 832, 1439, 1452, 1914, 2291, 2731, 3000, 3363, 3472, 5608, 5883, 6725, 6787, 7438, 8786, 10280, 11948, 12190, 13135, 15170, 15645, 22407, 26232, 27099, 32773, 33085, 40189, 40523, 48068, 51187
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OFFSET
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1,2
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COMMENTS
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These are the k-values of the lower envelope of the scatter band of the deviation of the binary weight of 3^k from half the length of the corresponding binary number. The corresponding negated differences are given in A372098.
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LINKS
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PROG
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(PARI) a372097(upto) = {my (dm=-oo); for (k=0, upto, my (p=3^k, h=hammingweight(p), b=#binary(p)/2, d=b-h); if (d>dm, print1(k, ", "); dm=d))};
a372097(60000)
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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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A372264
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a(n) = n! - n^2 + 2n - 1.
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1, 1, 2, 15, 104, 695, 5004, 40271, 362816, 3628719, 39916700, 479001479, 6227020656, 87178291031, 1307674367804, 20922789887775, 355687428095744, 6402373705727711, 121645100408831676, 2432902008176639639, 51090942171709439600, 1124000727777607679559, 25852016738884976639516
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OFFSET
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1,3
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COMMENTS
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The number of distinct cards in a deck that has each card twice to perform the n-card trick, where the assistant chooses the hidden card. This number corresponds to a particular strategy, where if there is a duplicate card, then the assistant puts one of those duplicates on the far left and hides the other. The assistant then arranges all the other cards in nondecreasing order. If there are no duplicates, the strategy is similar to the standard one.
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LINKS
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Michael Kleber and Ravi Vakil, The best card trick, The Mathematical Intelligencer 24 (2002), 9-11.
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MATHEMATICA
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Table[(n! - n^2 + 2 n - 1), {n, 1, 25}]
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PROG
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(Python)
from math import factorial
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CROSSREFS
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KEYWORD
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nonn,easy,new
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AUTHOR
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STATUS
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approved
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A370612
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The smallest number whose prime factor concatenation, when written in base n, does not contain 0 and contains all digits 1,...,(n-1) at least once.
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0
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3, 5, 14, 133, 706, 2490, 24258, 217230, 2992890, 24674730, 647850030
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OFFSET
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2,1
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COMMENTS
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All terms are squarefree. Many thanks to Michael Branicky for pointing out errors in the terms in the original submission.
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LINKS
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FORMULA
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EXAMPLE
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a(2) = 3 = 3 whose prime factors in base 2 is: 11.
a(3) = 5 = 5 whose prime factors in base 3 is: 12.
a(4) = 14 = 2*7 whose prime factors in base 4 is: 2, 13.
a(5) = 133 = 7*19 whose prime factors in base 5 is: 12, 34.
a(6) = 706 = 2*353 whose prime factors in base 6 is: 2, 1345.
a(7) = 2490 = 2*3*5*83 whose prime factors in base 7 is: 2, 3, 5, 146.
a(8) = 24258 = 2*3*13*311 whose prime factors in base 8 is: 2, 3, 15, 467.
a(9) = 217230 = 2*3*5*13*557 whose prime factors in base 9 is: 2, 3, 5, 14, 678.
a(10) = 2992890 = 2*3*5*67*1489.
a(11) = 24674730 = 2*3*5*19*73*593 whose prime factors in base 11 is: 2, 3, 5, 18, 67, 49a.
a(12) = 647850030 = 2*3*5*19*1136579 whose prime factors in base 12 is: 2, 3, 5, 17, 4698ab.
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PROG
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(Python)
from math import factorial
from itertools import count
from sympy import primefactors
from sympy.ntheory import digits
def A370612(n): return next(k for k in count(max(factorial(n-1), 2)) if 0 not in (s:=set.union(*(set(digits(p, n)[1:]) for p in primefactors(k)))) and len(s) == n-1)
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CROSSREFS
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KEYWORD
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nonn,base,more,new
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AUTHOR
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STATUS
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approved
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