A284877 Irregular triangle T(n,k) for 1 <= k <= n/2 + 1: T(n,k) = number of double palindrome structures of length n using exactly k different symbols.
1, 1, 1, 1, 3, 1, 7, 2, 1, 15, 5, 1, 25, 21, 3, 1, 49, 42, 7, 1, 79, 122, 44, 4, 1, 129, 225, 90, 9, 1, 211, 570, 375, 80, 5, 1, 341, 990, 715, 165, 11, 1, 517, 2321, 2487, 930, 132, 6, 1, 819, 3913, 4550, 1820, 273, 13, 1, 1275, 8827, 14350, 8330, 2009, 203, 7
Offset: 1
Examples
Triangle starts: 1 1 1 1 3 1 7 2 1 15 5 1 25 21 3 1 49 42 7 1 79 122 44 4 1 129 225 90 9 1 211 570 375 80 5 1 341 990 715 165 11 1 517 2321 2487 930 132 6 1 819 3913 4550 1820 273 13 1 1275 8827 14350 8330 2009 203 7 1 1863 14480 25515 15750 3990 420 15 1 2959 31802 75724 64004 23296 3920 296 8 1 4335 51425 132090 118167 44982 7854 612 17 1 6703 110928 376779 445275 229257 57078 7074 414 9 1 9709 177270 647995 807975 433713 111720 14250 855 19 1 15067 377722 1798175 2892470 2023135 698670 126300 12000 560 10 .... The first few structures are: n = 1: a => 1 n = 2: aa; ab => 1 + 1 n = 3: aaa; aab, aba, abb => 1 + 3 n = 4: aaaa; aaab, aaba, aabb, abaa, abab, abba, abbb; abac, abcb => 1 + 7 + 2
Links
- Andrew Howroyd, Table of n, a(n) for n = 1..960
Crossrefs
Programs
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Mathematica
r[d_, k_]:=If[OddQ[d], d*k^((d + 1)/2), (d/2)*(k + 1)*k^(d/2)]; a[n_, k_]:= r[n, k] - Sum[If[d
Indranil Ghosh, Apr 07 2017 *) -
PARI
r(d,k)=if (d % 2 == 0, (d/2)*(stirling(d/2,k,2)+stirling(d/2+1,k,2)), d*stirling((d+1)/2, k,2)); a(n,k) = r(n,k) - sumdiv(n,d, if (d
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Python
from sympy import totient, divisors, binomial, factorial def r(d, k): return (d//2)*(k + 1)*k**(d//2) if d%2 == 0 else d*k**((d + 1)//2) def a(n, k): return r(n, k) - sum([totient(n//d)*a(d, k) for d in divisors(n) if d
Indranil Ghosh, Apr 07 2017
Formula
T(n, k) = (Sum_{j=0..k} (-1)^j * binomial(k, j) * A284873(n, k-j)) / k!.
T(n, k) = r(n, k) - Sum_{d|n, d
Comments