A002412 -id:A002412 - cp's OEIS frontend

A338523 Triangle T(n,m) = (2mn+2n-2m^2+1)C(2n+2,2m+1)/(4n+2).

1, 2, 2, 3, 14, 3, 4, 44, 44, 4, 5, 100, 238, 100, 5, 6, 190, 828, 828, 190, 6, 7, 322, 2233, 4092, 2233, 322, 7, 8, 504, 5096, 14872, 14872, 5096, 504, 8, 9, 744, 10332, 43992, 70070, 43992, 10332, 744, 9, 10, 1050, 19176, 112200, 260780, 260780, 112200, 19176, 1050, 10

Offset: 0

Vladimir Kruchinin, Nov 01 2020

			1,
2, 2,
3, 14, 3,
4, 44, 44, 4,
5, 100, 238, 100, 5,
6, 190, 828, 828, 190, 6,
7, 322, 2233, 4092, 2233, 322, 7

Cf. A045623, A086645.

2nd column=2*A002412.

Table[Sum[Binomial[n + 1, 2 k + 1] Binomial[n - 2 k, m - k] (k + 1)*4^k, {k, 0, n} ], {n, 0, 9}, {m, 0, n}] // Flatten (* Michael De Vlieger, Nov 04 2020 *)

T(n,m):=((2*m*n+2*n-2*m^2+1)*binomial(2*n+2,2*m+1))/(4*n+2);

G.f.: (1/(1-x-x*y-4*x^2*y/(1-x-x*y)))^2.
T(n,m) = Sum_{k=0..n} C(n+1,2*k+1)*C(n-2*k,m-k)*(k+1)*4^k.
A045563(n) = (Sum_{m=0..n} T(n,m))/2^n.

A346083 Triangle, read by rows, defined by recurrence: T(n,k) = T(n-1,k-1) + (-1)^k * (2 * k + 1) * T(n-1,k) for 0 < k < n with initial values T(n,0) = T(n,n) = 1 for n >= 0 and T(i,j) = 0 if j < 0 or j > i.

Original entry on oeis.org

1, 1, 1, 1, -2, 1, 1, 7, 3, 1, 1, -20, 22, -4, 1, 1, 61, 90, 50, 5, 1, 1, -182, 511, -260, 95, -6, 1, 1, 547, 2373, 2331, 595, 161, 7, 1, 1, -1640, 12412, -13944, 7686, -1176, 252, -8, 1, 1, 4921, 60420, 110020, 55230, 20622, 2100, 372, 9, 1, 1, -14762, 307021, -709720, 607090, -171612, 47922, -3480, 525, -10, 1

Offset: 0

Werner Schulte, Jul 04 2021

			The triangle T(n,k) for 0 <= k <= n starts:
n\k :  0      1      2       3      4      5     6    7  8  9
=============================================================
  0 :  1
  1 :  1      1
  2 :  1     -2      1
  3 :  1      7      3       1
  4 :  1    -20     22      -4      1
  5 :  1     61     90      50      5      1
  6 :  1   -182    511    -260     95     -6     1
  7 :  1    547   2373    2331    595    161     7    1
  8 :  1  -1640  12412  -13944   7686  -1176   252   -8  1
  9 :  1   4921  60420  110020  55230  20622  2100  372  9  1
  etc.

Cf. A000012 (column 0 and main diagonal), A014983 (column 1), A181983 (1st subdiagonal), A002412 (2nd subdiagonal), A264851 (3rd subdiagonal without signs).

Cf. A051159.

from functools import cache
@cache
def T(n, k):
    if k == 0 or k == n: return 1
    return T(n-1, k-1) + (-1)**k*(2*k + 1)*T(n-1, k)
for n in range(10):
    print([T(n, k) for k in range(n+1)]) # Peter Luschny, Jul 22 2021

G.f. of column k >= 0: col(t,k) = Sum_{n >= k} T(n,k) * t^n = t^k / (Product_{i=0..k} (1 - (-1)^i * (2 * i + 1) * t)), i.e., col(t,k) = col(t,k-1) * t / (1 - (-1)^k * (2 * k + 1) * t) for k > 0.
Matrix inverse M = T^(-1) has row polynomials p(n,x) = Sum_{k=0..n} M(n,k) * x^k = Product_{i=1..n} (x + (-1)^i * (2 * i - 1)) for n >= 0 and empty product 1, i.e., p(n,x) = p(n-1,x) * (x + (-1)^n * (2 * n - 1)) for n > 0 with initial value p(0,x) = 1.
Conjecture: E.g.f. of column k >= 0: Sum_{n >= k} T(n,k) * t^n / (n!) = (Sum_{i=0..k} (-1)^(i * (i + 1 ) / 2) * binomial(k,floor((k - i) / 2)) * exp((-1)^i * (2 * i + 1) * t)) * (-1)^(k * (k - 1) / 2) / (4^k * (k!)), i.e., T(n,k) = (Sum_{i=0..k} (-1)^(i * (i + 1) / 2) * binomial(k,floor((k - i) / 2)) * ((-1)^i * (2 * i + 1))^n) * (-1)^(k * (k - 1) / 2) / (4^k * (k!)) for 0 <= k <= n.
Conjecture: E.g.f. of column k >= 0: Sum_{n >= k} T(n,k) * t^n / (n!) = exp(t) * (exp(4*t) - 1)^k / (4^k * (k!) * exp(4*t*floor((k+1)/2))), i.e., T(n,k) = (Sum_{i=0..k} (-1)^i * binomial(k,i) *(1 + 4*i - 4*floor((k+1)/2))^n) * (-1)^k / (4^k * (k!)) for 0 <= k <= n. Proved by Burkhard Hackmann and Werner Schulte (distinction of two cases: odd k, even k). - Werner Schulte, Aug 03 2021

A362888 a(1) = 1, a(n) = (3k + 1)(6k + 1)(8*k + 1), where k = Product_{i=1..n-1} a(i).

Original entry on oeis.org

1, 252, 2310152797, 28410981127871160285705816883937448685

Offset: 1

Ivan N. Ianakiev, May 08 2023

nonn

A sequence of pairwise relatively prime hexagonal pyramidal numbers. Its infinitude implies, by the Fundamental theorem of arithmetic, the infinitude of primes.
Building on an idea by Sierpinsky (see References): For m > 5, the general term of the sequence of m-gonal pyramidal numbers is a(n) = n*(n+1)*((m-2)*n - (m-5))/6. Therefore, for m > 5, there are infinitely many sequences of pairwise relatively prime m-gonal pyramidal numbers, with first term any positive m-gonal pyramidal
number and general term of the form a(n) = (3*k + 1)*(6*k + 1)*(2*k*(m - 2) + 1), where k = Product_{i=1..n-1} a(i). Corollary: There are infinitely many sequences of m-gonal pyramidal numbers to base the proof of the infinitude of primes on.

W. Sierpinski, 250 Problems in Elementary Number Theory. New York: American Elsevier, 1970. Problem #43.

Cf. A002412, A360826.

a[1]=1; a[n_]:=Module[{k=Product[a[i],{i,1,n-1}]},(3*k+1)*(6*k+1)*(8*k+1)];
a/@Range[5]

a(1) = 1, a(n) = (3*k + 1)*(6*k + 1)*(8*k + 1), where k = Product_{i=1..n-1} a(i).

cp's OEIS Frontend

A338523 Triangle T(n,m) = (2mn+2n-2m^2+1)C(2n+2,2m+1)/(4n+2).

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A346083 Triangle, read by rows, defined by recurrence: T(n,k) = T(n-1,k-1) + (-1)^k * (2 * k + 1) * T(n-1,k) for 0 < k < n with initial values T(n,0) = T(n,n) = 1 for n >= 0 and T(i,j) = 0 if j < 0 or j > i.

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Python

Formula

A362888 a(1) = 1, a(n) = (3k + 1)(6k + 1)(8*k + 1), where k = Product_{i=1..n-1} a(i).

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Formula

cp's OEIS Frontend

A338523 Triangle T(n,m) = (2*m*n+2*n-2*m^2+1)*C(2*n+2,2*m+1)/(4*n+2).

Views

Author

Keywords

Examples

Crossrefs

Programs

Mathematica

Maxima

Formula

A346083 Triangle, read by rows, defined by recurrence: T(n,k) = T(n-1,k-1) + (-1)^k * (2 * k + 1) * T(n-1,k) for 0 < k < n with initial values T(n,0) = T(n,n) = 1 for n >= 0 and T(i,j) = 0 if j < 0 or j > i.

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Author

Keywords

Examples

Crossrefs

Programs

Python

Formula

A362888 a(1) = 1, a(n) = (3*k + 1)*(6*k + 1)*(8*k + 1), where k = Product_{i=1..n-1} a(i).

Views

Author

Keywords

Comments

References

Crossrefs

Programs

Mathematica

Formula

A338523 Triangle T(n,m) = (2mn+2n-2m^2+1)C(2n+2,2m+1)/(4n+2).

A362888 a(1) = 1, a(n) = (3k + 1)(6k + 1)(8*k + 1), where k = Product_{i=1..n-1} a(i).