Matthew C. Russell - cp's OEIS frontend

A296010 Sum of the squares of the number of parts in all partitions of n.

0, 1, 5, 14, 34, 68, 133, 232, 402, 652, 1048, 1609, 2465, 3640, 5358, 7694, 10993, 15399, 21498, 29520, 40394, 54572, 73425, 97756, 129710, 170525, 223428, 290552, 376551, 484819, 622317, 794167, 1010515, 1279376, 1615126, 2029948, 2544600, 3176856, 3956277

Offset: 0

Matthew C. Russell, Dec 02 2017

nonn

			For n=4, the 5 partitions of 4 are 4, 3+1, 2+2, 2+1+1, and 1+1+1+1. These have 1, 2, 2, 3, and 4 parts, respectively. The sum of the squares is 1+4+4+9+16=34.

Charles R Greathouse IV, Table of n, a(n) for n = 0..10000 (terms 0..115 from Robert G. Wilson v)

Cf. A006128, A008284, A036037.

K:=[]:
for n from 0 to 20 do
co:=0:
for L in combinat[partition](n) do
co:=co+nops(L)^2:
od:
K:=[op(K),co]:
od:
K;
# second Maple program:
b:= proc(n, i, c) option remember; `if`(n=0 or i=1,
      (n+c)^2, `if`(i>n, 0, b(n-i, i, c+1))+b(n, i-1, c))
    end:
a:= n-> b(n$2, 0):
seq(a(n), n=0..50);  # Alois P. Heinz, Dec 02 2017

f[n_] := Sum[i^2 (Length@ IntegerPartitions[n, {i}]), {i, n}]; Array[f, 34, 0] (* Robert G. Wilson v, Dec 02 2017 *)
b[n_, i_, c_] := b[n, i, c] = If[n == 0 || i == 1,
     (n + c)^2, If[i > n, 0, b[n - i, i, c + 1]] + b[n, i - 1, c]];
a[n_] := b[n, n, 0];
a /@ Range[0, 50] (* Jean-François Alcover, Jun 06 2021, after Alois P. Heinz *)

first(n)=my(x='x+O('x^(n+1)),pr=1); concat(0,Vec(sum(j=1,n,pr*=1-x^j; j^2*x^j/pr))) \\ Charles R Greathouse IV, Dec 02 2017

G.f.: Sum_{j>=1} j^2*x^j / Product_{i=1..j} (1-x^i). - Alois P. Heinz, Dec 02 2017

A272397 Number of partitions of n into parts congruent to 1, 3, 6, 8 (mod 9).

Original entry on oeis.org

1, 1, 1, 2, 2, 2, 4, 4, 5, 7, 8, 9, 13, 14, 16, 21, 24, 27, 35, 39, 45, 55, 62, 70, 86, 96, 109, 130, 146, 164, 195, 217, 245, 285, 319, 357, 415, 461, 517, 592, 660, 735, 840, 931, 1038, 1175, 1304, 1446, 1634, 1805, 2002, 2246, 2482, 2742, 3070, 3381, 3734

Offset: 0

Matthew C. Russell, Apr 28 2016

nonn

"Sum side" conjecture: also equals number of partitions pi = (pi_1, pi_2, ...) of n (with pi_1 >= pi_2 >= ...) such that pi(i)-pi(i+2) >= 3 and, if pi(i) - pi(i+1) <= 1, then pi(i) + pi(i+1) is congruent to 0 (mod 3).

			For n=10, the a(10)=8 partitions are 10, 8+1+1, 6+3+1, 6+1+1+1, 3+3+3+1, 3+3+1+1+1+1. 3+1+1+1+1+1+1+1, and 1+1+1+1+1+1+1+1+1+1.
For the conjectured "sum side", the a(10)=8 partitions are 10, 9+1, 8+2, 7+3, 7+2+1, 6+4, 6+3+1, and 5+4+1.

S. Kanade and M. C. Russell, IdentityFinder and some new identities of Rogers-Ramanujan type, Exp. Math. 24:4 (2015), pp. 419-423.

Cf. A000726: partitions of 3n into parts == {3,6} mod 9.

Table[Length@ Select[IntegerPartitions@ n, AllTrue[Mod[#, 9], MemberQ[{1, 3, 6, 8}, #] &] &], {n, 0, 50}] (* Michael De Vlieger, Apr 28 2016, Version 10 *)

A230058 Numbers of the form k + wt(k) for at least two distinct k, where wt(k) = A000120(k) is the binary weight of k.

Original entry on oeis.org

5, 14, 17, 19, 22, 31, 33, 36, 38, 47, 50, 52, 55, 64, 67, 70, 79, 82, 84, 87, 96, 98, 101, 103, 112, 115, 117, 120, 129, 131, 132, 134, 143, 146, 148, 151, 160, 162, 165, 167, 176, 179, 181, 184, 193, 196, 199, 208, 211, 213, 216, 225, 227, 230, 232, 241, 244

Offset: 1

Matthew C. Russell, Oct 07 2013

The positions of entries greater than 1 in A228085, or numbers that appear multiple times in A092391.

Numbers that can be expressed as the sum of distinct terms of the form 2^n+1, n=0,1,... in multiple ways.

			5 = 3 + 2 = 4 + 1, so 5 is in this list.

Vincenzo Librandi and Donovan Johnson, Table of n, a(n) for n = 1..10000 (first 1000 terms from Vincenzo Librandi)
Index entries for Colombian or self numbers and related sequences

Cf. A000120, A010061, A228088, A230091, A230092.

Sort[Transpose[Select[Tally[Table[k + Total[IntegerDigits[k, 2]], {k, 0, 300}]], #[[2]] > 1 &]][[1]]] (* T. D. Noe, Oct 09 2013 *)

A229744 Values of records in A096303.

Original entry on oeis.org

0, 1, 4, 6, 7, 10, 11, 20, 21, 23, 32, 33, 38, 47, 51, 52, 53, 54, 78, 101, 106, 122, 177, 205, 216, 370, 373, 383, 393, 411, 753, 757, 758, 763

Offset: 1

Matthew C. Russell and Nathan Fox, Oct 06 2013

Index entries for Colombian or self numbers and related sequences

Cf. A010062, A096303, A229743.

a(15)-a(34) from Donovan Johnson, Oct 07 2013

A229743 Positions of records in A096303.

Original entry on oeis.org

1, 4, 6, 21, 46, 86, 359, 914, 1432, 1682, 1810, 1816, 2840, 3608, 15141, 15909, 24120, 28197, 30245, 31781, 16514098, 66058853, 67105381, 117437029, 134148709, 134210149, 234873445, 251650661, 267903589, 268411493, 268419685, 402637413, 532660837, 534757989

Offset: 1

Matthew C. Russell and Nathan Fox, Oct 06 2013

Index entries for Colombian or self numbers and related sequences

Cf. A010062, A096303, A229744.

a(15)-a(34) from Donovan Johnson, Oct 07 2013

A211882 Number of integral circulant graphs with perfect state transfer on 4*n vertices.

Original entry on oeis.org

2, 2, 4, 4, 4, 16, 4, 8, 8, 16, 4, 256, 4, 16, 16, 16, 4, 512, 4, 256, 16, 16, 4, 4096, 8, 16, 16, 256, 4, 65536, 4, 32, 16, 16, 16, 262144, 4, 16, 16, 4096, 4, 65536, 4, 256, 64, 16, 4, 65536, 8, 512, 16, 256, 4, 65536, 16, 4096, 16, 16, 4, 4294967296

Offset: 1

Matthew C. Russell, Apr 25 2012

nonn

			For n=1 there are a(1)=2 graphs with perfect state transfer: C_4 and the disjoint union of two copies of K_2.

Fomin, Sergey; Zelevinsky, Andrei; The Laurent phenomenon, arXiv:math/0104241v1 [math.CO] (2001), Advances in Applied Mathematics 28 (2002), 119-144.

A208222 a(n) = (a(n-1)^3*a(n-3)^2+a(n-2))/a(n-4) with a(0)=a(1)=a(2)=a(3)=1.

Original entry on oeis.org

1, 1, 1, 1, 2, 9, 731, 1562471573, 154486807085783774292345385804

Offset: 0

Matthew C. Russell, Apr 25 2012

nonn

This is the case a=2, b=1, c=3, y(0)=y(1)=y(2)=y(3)=1 of the recurrence shown in the Example 3.3 of "The Laurent phenomenon" (see Link lines, p. 10).

Seiichi Manyama, Table of n, a(n) for n = 0..11
Sergey Fomin and Andrei Zelevinsky, The Laurent phenomenon, arXiv:math/0104241v1 [math.CO] (2001), Advances in Applied Mathematics 28 (2002), 119-144.

Cf. A048736, A208219, A208221, A208225.

y:=proc(n) if n<4 then return 1: fi: return (y(n-1)^3*y(n-3)^2+y(n-2))/y(n-4): end:
seq(y(n),n=0..9);

a[n_]:=If[n<4,1, (a[n - 1]^3*a[n - 3]^2 + a[n - 2])/a[n - 4]]; Table[a[n], {n, 0, 11}] (* Indranil Ghosh, Mar 19 2017 *)
nxt[{a_,b_,c_,d_}]:={b,c,d,(d^3 b^2+c)/a}; NestList[nxt,{1,1,1,1},10][[All,1]] (* Harvey P. Dale, May 31 2020 *)

A208221 a(n)=(a(n-1)^2*a(n-3)^2+a(n-2))/a(n-4) with a(0)=a(1)=a(2)=a(3)=1.

Original entry on oeis.org

1, 1, 1, 1, 2, 5, 27, 2921, 106653026, 1658455747832683945, 869174798276372512100586962107665002957113

Offset: 0

Matthew C. Russell, Apr 25 2012

nonn

This is the case a=2, b=1, c=2, y(0)=y(1)=y(2)=y(3)=1 of the recurrence shown in the Example 3.3 of "The Laurent phenomenon" (see Link lines, p. 10).

The next term (a(11)) has 97 digits. - Harvey P. Dale, Dec 17 2017

Seiichi Manyama, Table of n, a(n) for n = 0..13
Sergey Fomin and Andrei Zelevinsky, The Laurent phenomenon, arXiv:math/0104241v1 [math.CO] (2001); Advances in Applied Mathematics 28 (2002), 119-144.

Cf. A048736, A208218, A208220, A208222, A208224.

y:=proc(n) if n<4 then return 1: fi: return (y(n-1)^2*y(n-3)^2+y(n-2))/y(n-4): end:
seq(y(n),n=0..11);

a[n_] := a[n] = If[n <= 3, 1, (a[n-1]^2*a[n-3]^2 + a[n-2])/a[n-4]];
Table[a[n], {n, 0, 13}] (* Jean-François Alcover, Nov 24 2017 *)
RecurrenceTable[{a[0]==a[1]==a[2]==a[3]==1,a[n]==(a[n-1]^2 a[n-3]^2+ a[n-2])/ a[n-4]},a,{n,12}] (* Harvey P. Dale, Dec 17 2017 *)

A208219 a(n)=(a(n-1)^3*a(n-3)+a(n-2))/a(n-4) with a(0)=a(1)=a(2)=a(3)=1.

Original entry on oeis.org

1, 1, 1, 1, 2, 9, 731, 781235791, 2145650135491172007486084385, 802327342392981520933850619811649523436811893002103478524225246677189521545661182074

Offset: 0

Matthew C. Russell, Apr 25 2012

nonn

This is the case a=1, b=1, c=3, y(0)=y(1)=y(2)=y(3)=1 of the recurrence shown in the Example 3.3 of "The Laurent phenomenon" (see Link lines, p. 10).

The next term (a(10)) has 258 digits. - Harvey P. Dale, Sep 21 2016

Seiichi Manyama, Table of n, a(n) for n = 0..11
Sergey Fomin and Andrei Zelevinsky, The Laurent phenomenon, arXiv:math/0104241v1 [math.CO] (2001), Advances in Applied Mathematics 28 (2002), 119-144.

Cf. A048736, A208218, A208222.

y:=proc(n) if n<4 then return 1: fi: return (y(n-1)^3*y(n-3)+y(n-2))/y(n-4): end:
seq(y(n),n=0..9);

RecurrenceTable[{a[0]==a[1]==a[2]==a[3]==1,a[n]==(a[n-1]^3 a[n-3]+ a[n-2])/ a[n-4]},a,{n,10}] (* Harvey P. Dale, Sep 21 2016 *)

A208224 a(n)=(a(n-1)^2*a(n-3)^3+a(n-2))/a(n-4) with a(0)=a(1)=a(2)=a(3)=1.

Original entry on oeis.org

1, 1, 1, 1, 2, 5, 27, 5837, 2129410576, 17850077316687753782569, 2346851008195218976646246398770505953580095510848345967

Offset: 0

Matthew C. Russell, Apr 25 2012

nonn

This is the case a=3, b=1, c=2, y(0)=y(1)=y(2)=y(3)=1 of the recurrence shown in the Example 3.3 of "The Laurent phenomenon" (see Link lines, p. 10).

The next term (a(11)) has 133 digits. - Harvey P. Dale, Mar 06 2017

Seiichi Manyama, Table of n, a(n) for n = 0..13
Sergey Fomin and Andrei Zelevinsky, The Laurent phenomenon, arXiv:math/0104241v1 [math.CO] (2001), Advances in Applied Mathematics 28 (2002), 119-144.

Cf. A048736, A208221, A208223, A208225, A208227.

y:=proc(n) if n<4 then return 1: fi: return (y(n-1)^2*y(n-3)^3+y(n-2))/y(n-4): end:
seq(y(n),n=0..11);

RecurrenceTable[{a[0]==a[1]==a[2]==a[3]==1,a[n]==(a[n-1]^2*a[n-3]^3+ a[n-2])/ a[n-4]},a,{n,10}] (* Harvey P. Dale, Mar 06 2017 *)

cp's OEIS Frontend

User: Matthew C. Russell

A296010 Sum of the squares of the number of parts in all partitions of n.

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A272397 Number of partitions of n into parts congruent to 1, 3, 6, 8 (mod 9).

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A230058 Numbers of the form k + wt(k) for at least two distinct k, where wt(k) = A000120(k) is the binary weight of k.

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A229744 Values of records in A096303.

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A229743 Positions of records in A096303.

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Extensions

A211882 Number of integral circulant graphs with perfect state transfer on 4*n vertices.

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A208222 a(n) = (a(n-1)^3*a(n-3)^2+a(n-2))/a(n-4) with a(0)=a(1)=a(2)=a(3)=1.

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A208221 a(n)=(a(n-1)^2*a(n-3)^2+a(n-2))/a(n-4) with a(0)=a(1)=a(2)=a(3)=1.

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Maple

Mathematica

A208219 a(n)=(a(n-1)^3*a(n-3)+a(n-2))/a(n-4) with a(0)=a(1)=a(2)=a(3)=1.

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