A154323 -id:A154323 - cp's OEIS frontend

A360693 Number T(n,k) of sets of n words of length n over binary alphabet where the first letter occurs k times; triangle T(n,k), n>=0, n-signum(n)<=k<=n*(n-1)+signum(n), read by rows.

1, 1, 1, 2, 2, 2, 3, 10, 15, 15, 10, 3, 4, 37, 108, 228, 336, 394, 336, 228, 108, 37, 4, 5, 101, 600, 2150, 5645, 11680, 19752, 27820, 32935, 32935, 27820, 19752, 11680, 5645, 2150, 600, 101, 5, 6, 226, 2490, 14745, 61770, 200529, 535674, 1211485, 2368200

Offset: 0

Alois P. Heinz, Feb 16 2023

T(n,k) is defined for all n >= 0 and k >= 0. The triangle contains only the positive elements.

			T(2,3) = 2: {aa,ab}, {aa,ba}.
T(3,3) = 10: {aab,abb,bbb}, {aab,bab,bbb}, {aab,bba,bbb}, {aba,abb,bbb}, {aba,bab,bbb}, {aba,bba,bbb}, {abb,baa,bbb}, {abb,bab,bba}, {baa,bab,bbb}, {baa,bba,bbb}.
T(4,3) = 4: {abbb,babb,bbab,bbbb}, {abbb,babb,bbba,bbbb}, {abbb,bbab,bbba,bbbb}, {babb,bbab,bbba,bbbb}.
Triangle T(n,k) begins:
  1;
  1, 1;
  .  2, 2,  2;
  .  .  3, 10, 15,  15,  10,    3;
  .  .  .   4, 37, 108, 228,  336,  394,   336,   228,   108,    37,     4;
  .  .  .   .   5, 101, 600, 2150, 5645, 11680, 19752, 27820, 32935, 32935, ...;
  ...

Alois P. Heinz, Rows n = 0..33, flattened

Row sums give A014070.
Column sums give A360695.
Main diagonal T(n,n) gives A154323(n-1) for n>=1.
T(n,n-1) gives A000027(n) for n>=1.
T(2n,2n^2) gives A360702.
Cf. A000290, A057427, A220886 (similar triangle for multisets).

g:= proc(n, i, j) option remember; expand(`if`(j=0, 1, `if`(i<0, 0, add(
      g(n, i-1, j-k)*x^(i*k)*binomial(binomial(n, i), k), k=0..j))))
    end:
T:= n-> (p-> seq(coeff(p, x, i), i=n-signum(n)..n*(n-1)+signum(n)))(g(n$3)):
seq(T(n), n=0..6);

g[n_, i_, j_] := g[n, i, j] = Expand[If[j == 0, 1, If[i < 0, 0, Sum[g[n, i - 1, j - k]*x^(i*k)*Binomial[Binomial[n, i], k], {k, 0, j}]]]];
T[n_] := Table[Coefficient[#, x, i], {i, n - Sign[n], n(n - 1) + Sign[n]}]&[g[n, n, n]];
Table[T[n], {n, 0, 6}] // Flatten (* Jean-François Alcover, May 26 2023, after Alois P. Heinz *)

T(n,k) = T(n,n^2-k).

A113582 Triangle T(n,m) read by rows: T(n,m) = (n - m)(n - m + 1)m*(m + 1)/4 + 1.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 4, 4, 1, 1, 7, 10, 7, 1, 1, 11, 19, 19, 11, 1, 1, 16, 31, 37, 31, 16, 1, 1, 22, 46, 61, 61, 46, 22, 1, 1, 29, 64, 91, 101, 91, 64, 29, 1, 1, 37, 85, 127, 151, 151, 127, 85, 37, 1, 1, 46, 109, 169, 211, 226, 211, 169, 109, 46, 1

Offset: 1

Roger L. Bagula, Aug 25 2008

From Paul Barry, Jan 07 2009: (Start)
This triangle follows a general construction method as follows: Let a(n) be an integer sequence with a(0)=1, a(1)=1. Then T(n,k,r) := [k<=n](1+r*a(k)*a(n-k)) defines a symmetrical triangle.
Row sums are n + 1 + r*Sum_{k=0..n} a(k)*a(n-k) and central coefficients are 1+r*a(n)^2.
Here a(n) = C(n+1,2) and r=1.
Row sums are A154322 and central coefficients are A154323. (End)

			{1},
{1, 1},
{1, 2, 1},
{1, 4, 4, 1},
{1, 7, 10, 7, 1},
{1, 11, 19, 19, 11, 1},
{1, 16, 31, 37, 31, 16, 1},
{1, 22, 46, 61, 61, 46, 22, 1},
{1, 29, 64, 91, 101, 91, 64, 29, 1},
{1, 37, 85, 127, 151, 151, 127, 85, 37, 1},
{1, 46, 109, 169, 211, 226, 211, 169, 109, 46, 1}

G. C. Greubel, Rows n=0..100 of triangle, flattened

/* As triangle: */ [[(n-m)*(n-m+1)*m*(m+1)/4+1: m in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Sep 12 2016

t[n_, m_] = (n - m)*(n - m + 1)*m*(m + 1)/4 + 1; Table[Table[t[n, m], {m, 0, n}], {n, 0, 10}]//Flatten

for(n=0,15, for(k=0,n, print1((n-k)*(n-k+1)*k*(k+1)/4 + 1, ", "))) \\ G. C. Greubel, Aug 31 2018

T(n,m) = (n - m)*(n - m + 1)*m*(m + 1)/4 + 1.

A360660 Number of inequivalent n X n {0,1} matrices modulo permutation of the rows, with exactly n 1's.

Original entry on oeis.org

1, 1, 4, 20, 133, 1027, 9259, 94033, 1062814, 13176444, 177427145, 2573224238, 39924120823, 658921572675, 11513293227271, 212109149134617, 4105637511110979, 83238756058333277, 1762856698153603049, 38905470655863251479, 892840913430059075405

Offset: 0

Alois P. Heinz, Feb 15 2023

nonn

Also the number of multisets of n words of length n over binary alphabet where the first letter occurs n times. E.g., a(2) = 4: {aa,bb}, {ab,ab}, {ab,ba}, {ba,ba}.

			a(3) = 20: [111/000/000], [110/100/000], [110/010/000], [110/001/000], [101/100/000], [101/010/000], [101/001/000], [100/100/100], [100/100/010], [100/100/001], [100/011/000], [100/010/010], [100/010/001], [100/001/001], [011/010/000], [011/001/000], [010/010/010], [010/010/001], [010/001/001], [001/001/001].

Alois P. Heinz, Table of n, a(n) for n = 0..200

Main diagonal of A220886.

Cf. A091058, A091059, A154323, A246107, A256070, A360664, A383073.

g:= proc(n, i, j) option remember; expand(`if`(j=0, 1, `if`(i<0, 0, add(
      g(n, i-1, j-k)*x^(i*k)*binomial(binomial(n, i)+k-1, k), k=0..j))))
    end:
a:= n-> coeff(g(n$3), x, n):
seq(a(n), n=0..20);

g[n_, i_, j_] := g[n, i, j] = Expand[If[j == 0, 1, If[i < 0, 0, Sum[g[n, i - 1, j - k]*x^(i*k)*Binomial[Binomial[n, i] + k - 1, k], {k, 0, j}]]]];
a[n_] := SeriesCoefficient[g[n, n, n], {x, 0, n}];
Table[a[n], {n, 0, 20}] (* Jean-François Alcover, May 28 2023, after Alois P. Heinz *)
Table[SeriesCoefficient[Product[1/(1 - x^k)^Binomial[n, k], {k, 1, n}], {x, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Apr 15 2025 *)

a(n) = A220886(n,n).

a(n) = [x^n] Product_{k=1..n} 1/(1 - x^k)^binomial(n,k). - Vaclav Kotesovec, Apr 15 2025

A262182 Prime numbers of the form (n*(n+1)/2)^2 + 1.

Original entry on oeis.org

2, 37, 101, 1297, 4357, 14401, 44101, 90001, 164837, 246017, 608401, 894917, 1382977, 4326401, 8122501, 8561477, 9985601, 10497601, 38638657, 46049797, 52707601, 84272401, 121572677, 146168101, 165894401, 201526417, 259532101, 289680401, 404010001, 428738437

Offset: 1

Jean C. Lambry, Oct 02 2015

nonn

Sum_{n>=1} 1/a(n) = 0.538046187...

Chai Wah Wu, Table of n, a(n) for n = 1..5231

Cf. A154323, A217755.

Select[Accumulate[Range[250]]^2+1,PrimeQ] (* Harvey P. Dale, Nov 14 2024 *)

for(n=1, 1e3, if(isprime(k = (n*(n+1)/2)^2+1), print1(k", "))) \\ Altug Alkan, Oct 02 2015

def prime(n):
    if n == 1: return True
    d = n + 1
    c = n - 1
    while c > 0 and d % c:
        d += n
        c -= 1
    return bool(c == 1)
n = 1
i = 1
while i <= 500:
    target = (i * (i + 1)) // 2
    if prime(target):
        print(n, target*target+1)
        n += 1
    i += 1
# Jean C. Lambry, Oct 06 2015

a(n) = A154323(A217755(n)). - Michel Marcus, Oct 02 2015

a(n) = ((A217755(n)^2 + A217755(n))/2)^2 + 1. - Jean C. Lambry, Oct 07 2015

More terms from Altug Alkan, Oct 02 2015

A263689 a(n) = (2n^6 - 6n^5 + 5*n^4 - n^2 + 12)/12.

Original entry on oeis.org

1, 1, 2, 34, 277, 1301, 4426, 12202, 29009, 61777, 120826, 220826, 381877, 630709, 1002002, 1539826, 2299201, 3347777, 4767634, 6657202, 9133301, 12333301, 16417402, 21571034, 28007377, 35970001, 45735626, 57617002, 71965909, 89176277, 109687426, 133987426, 162616577, 196171009, 235306402, 280741826

Offset: 0

Ilya Gutkovskiy, Nov 20 2015

			a(0) = 1,
a(1) = 0^5 + 1 = 1,
a(2) = 1^5 + 1 = 2,
a(3) = 2^5 + 2 = 34,
a(4) = 3^5 + 34 = 227,
a(5) = 4^5 + 227 = 1301, etc.

Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,1).

Cf. A000124, A000539, A000584, A056520, A154323.

Table[(1/12) (12 + (-1 + n)^2 n^2 (-1 + 2 (-1 + n) n)), {n, 0, 35}]

first(m)=vector(m,n,n--;(2*n^6 - 6*n^5 + 5*n^4 - n^2 + 12)/12) \\ Anders Hellström, Nov 20 2015

G.f.: (1 - 6*x + 16*x^2 + 6*x^3 + 81*x^4 + 20*x^5 + 2*x^6)/(1 - x)^7.
a(n + 1) = a(n) + n^5, a(0) = 1.
a(n + 1) - a(n) = A000584(n).
a(n + 1) = A000539(n) + 1.
Sum_{n>0} 1/(a(n + 1) - a(n)) = zeta(5) = 1.036927755...

A267691 a(n) = (n + 1)(6n^4 - 21n^3 + 31n^2 - 31*n + 30)/30.

Original entry on oeis.org

1, 1, 2, 18, 99, 355, 980, 2276, 4677, 8773, 15334, 25334, 39975, 60711, 89272, 127688, 178313, 243849, 327370, 432346, 562667, 722667, 917148, 1151404, 1431245, 1763021, 2153646, 2610622, 3142063, 3756719, 4464000, 5274000, 6197521, 7246097, 8432018, 9768354

Offset: 0

Ilya Gutkovskiy, Jan 19 2016

			a(0) = 1,
a(1) = 1 + 0^4 = 1,
a(2) = 1 + 1^4 = 2,
a(3) = 2 + 2^4 = 18,
a(4) = 18+ 3^4 = 99, etc.

Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1)

Essentially the same as A000538.
Cf. A000124, A000583, A008514, A056520, A154323, A263689.
Cf. A013662 (zeta(4)).

[(n+1)*(6*n^4-21*n^3+31*n^2-31*n+30)/30: n in [0..35]]; // Vincenzo Librandi, Jan 20 2016

Table[(n + 1) (6 n^4 - 21 n^3 + 31 n^2 - 31 n + 30)/30, {n, 0, 30}]
LinearRecurrence[{6, -15, 20, -15, 6, -1}, {1, 1, 2, 18, 99, 355}, 40] (* Vincenzo Librandi, Jan 20 2016 *)

a(n)=(n+1)*(6*n^4-21*n^3+31*n^2-31*n+30)/30 \\ Charles R Greathouse IV, Jan 19 2016

Vec((1-5*x+11*x^2+x^3+16*x^4)/(x-1)^6 + O(x^100)) \\ Altug Alkan, Jan 19 2016

G.f.: (1 - 5*x + 11*x^2 + x^3 + 16*x^4)/(1 - x)^6.
a(n + 1) = a(n) + n^4.
a(n + 1) = A000538(n) + 1.
a(n + 2) - a(n) = A008514(n).
Sum_{n>=0} 1/a(n) = 2.570450909491318975...
Sum_{n>=1} 1/(a(n + 1) - a(n)) = zeta(4) = Pi^4/90.

cp's OEIS Frontend

A360693 Number T(n,k) of sets of n words of length n over binary alphabet where the first letter occurs k times; triangle T(n,k), n>=0, n-signum(n)<=k<=n*(n-1)+signum(n), read by rows.

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A113582 Triangle T(n,m) read by rows: T(n,m) = (n - m)*(n - m + 1)*m*(m + 1)/4 + 1.

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A360660 Number of inequivalent n X n {0,1} matrices modulo permutation of the rows, with exactly n 1's.

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A262182 Prime numbers of the form (n*(n+1)/2)^2 + 1.

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A263689 a(n) = (2*n^6 - 6*n^5 + 5*n^4 - n^2 + 12)/12.

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A267691 a(n) = (n + 1)*(6*n^4 - 21*n^3 + 31*n^2 - 31*n + 30)/30.

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Magma

Mathematica

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PARI

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A113582 Triangle T(n,m) read by rows: T(n,m) = (n - m)(n - m + 1)m*(m + 1)/4 + 1.

A263689 a(n) = (2n^6 - 6n^5 + 5*n^4 - n^2 + 12)/12.

A267691 a(n) = (n + 1)(6n^4 - 21n^3 + 31n^2 - 31*n + 30)/30.