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A269044
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a(n) = 13*n + 7.
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10
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7, 20, 33, 46, 59, 72, 85, 98, 111, 124, 137, 150, 163, 176, 189, 202, 215, 228, 241, 254, 267, 280, 293, 306, 319, 332, 345, 358, 371, 384, 397, 410, 423, 436, 449, 462, 475, 488, 501, 514, 527, 540, 553, 566, 579, 592, 605, 618, 631, 644, 657, 670, 683, 696, 709, 722, 735
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OFFSET
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0,1
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COMMENTS
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After 7 (which corresponds to n=0), all terms belong to A090767 because a(n) = 3*n*2*1 + 2*(n*2+2*1+n*1) + (n+2+1).
This sequence is related to A152741 by the recurrence A152741(n+1) = (n+1)*a(n+1) - Sum_{k = 0..n} a(k).
Any square mod 13 is one of 0, 1, 3, 4, 9, 10 or 12 (A010376) but not 7, and for this reason there are no squares in the sequence. Likewise, any cube mod 13 is one of 0, 1, 5, 8 or 12, therefore no a(k) is a cube.
The sum of the squares of any two terms of the sequence is also a term of the sequence, that is: a(h)^2 + a(k)^2 = a(h*(13*h+14) + k*(13*k+14) + 7). Therefore: a(h)^2 + a(k)^2 > a(a( h*(h+1) + k*(k+1) )) for h+k > 0.
The primes of the sequence are listed in A140371.
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LINKS
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Bruno Berselli, A description of the recursive method shown in the third comment: website Matem@ticamente (in Italian), 2008.
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FORMULA
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G.f.: (7 + 6*x)/(1 - x)^2.
Sum_{i=h..h+13*k} a(i) = a(h*(13*k + 1) + k*(169*k + 27)/2).
Sum_{i>=0} 1/a(i)^2 = 0.0257568950542502716970... = polygamma(1, 7/13)/13^2.
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MATHEMATICA
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13 Range[0, 60] + 7 (* or *) Range[7, 800, 13] (* or *) Table[13 n + 7, {n, 0, 60}]
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PROG
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(Magma) [13*n+7: n in [0..60]];
(Maxima) makelist(13*n+7, n, 0, 60);
(PARI) vector(60, n, n--; 13*n+7)
(Sage) [13*n+7 for n in (0..60)]
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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