A002411 -id:A002411 - cp's OEIS frontend

A136360 Square roots of the perfect squares in A133459.

6, 9, 12, 17, 22, 24, 25, 26, 60, 86, 99, 120, 188, 200, 202, 210, 214, 238, 243, 268, 336, 348, 415, 476, 481, 504, 524, 539, 565, 602, 693, 704, 720, 726, 732, 846, 899, 961, 965, 990, 1026, 1202, 1218, 1221, 1224, 1320, 1551, 1602, 1687, 1716, 1724, 1734

Offset: 1

Alexander Adamchuk, Dec 25 2007

nonn

Corresponding squares in A133459 are listed in A136359(n) = a(n)^2.

Note that some numbers in a(n) are also perfect squares: m = k^2 = {9, 25, 961, 17424, ...}. The corresponding numbers k such that a(n) = k^2 are listed in A136361.

			A133459 begins {2, 7, 12, 19, 24, 36, 41, 46, 58, 76, 80, 81, 93, 115, 127, 132, 144, 150, 166, 197, 201, 202, 214, 236, 252, 271, 289, ...}.
Thus a(1) = sqrt(36) = 6, a(2) = sqrt(81) = 9, a(3) = sqrt(144) = 12, a(4) = sqrt(289) = 17 that are the square roots of the perfect squares in A133459.

Cf. A136359, A136361, A133459, A002311, A002411, A053721.

Sqrt[ Select[ Intersection[ Flatten[ Table[ i^2*(i+1)/2 + j^2*(j+1)/2, {i,1,300}, {j,1,i} ] ] ], IntegerQ[ Sqrt[ # ] ] & ] ]

a(n) = sqrt(A136359(n)).

A136361 Square roots of the perfect squares in A136360; or numbers k such that k^4 is in A133459 = the sums of two nonzero pentagonal pyramidal numbers.

Original entry on oeis.org

3, 5, 31, 132, 1068, 9672, 50664, 145060

Offset: 1

Alexander Adamchuk, Dec 25 2007

Corresponding perfect squares in A136360 are a(n)^2 = {9, 25, 961, 17424, ...}. They correspond to the perfect fourth powers in A133459 = Sums of two nonzero pentagonal pyramidal numbers. a(n)^4 are the terms of A133459: {81, 525, 923521, 303595776, ...}. Note that the first three terms are prime.

a(9) > (5*10^20)^(1/4). - Donovan Johnson, Jun 12 2011

Cf. A136359, A136360, A133459, A002311, A002411, A053721.

Name corrected and a(5)-a(6) from Donovan Johnson, Nov 20 2010

a(7)-a(8) from Donovan Johnson, Jun 12 2011

A189073 Triangle read by rows: T(n,k) is the number of inversions in k-compositions of n for n >= 3, 2 <= k <= n-1.

Original entry on oeis.org

1, 1, 3, 2, 6, 6, 2, 12, 18, 10, 3, 18, 42, 40, 15, 3, 27, 78, 110, 75, 21, 4, 36, 132, 240, 240, 126, 28, 4, 48, 204, 460, 600, 462, 196, 36, 5, 60, 300, 800, 1290, 1302, 812, 288, 45, 5, 75, 420, 1300, 2490, 3108, 2548, 1332, 405, 55, 6, 90, 570, 2000, 4440, 6594, 6692, 4608, 2070, 550, 66

Offset: 3

N. J. A. Sloane, Apr 16 2011

The Heibach et al. reference has a table for n <= 14.

			Triangle begins:
1;
1,  3;
2,  6,   6;
2, 12,  18,   10;
3, 18,  42,   40,   15;
3, 27,  78,  110,   75,   21;
4, 36, 132,  240,  240,  126,   28;
4, 48, 204,  460,  600,  462,  196,   36;
5, 60, 300,  800, 1290, 1302,  812,  288,   45;
5, 75, 420, 1300, 2490, 3108, 2548, 1332,  405,  55;
6, 90, 570, 2000, 4440, 6594, 6692, 4608, 2070, 550, 66;
...
T(5,3) = 6 because we have: 3+1+1, 1+3+1, 1+1+3, 2+2+1, 2+1+2, 1+2+2 having 2,1,0,2,1,0 inversions respectively. - _Geoffrey Critzer_, Mar 19 2014

Alois P. Heinz, Rows n = 3..143, flattened
S. Heubach, A. Knopfmacher, M. E. Mays and A. Munagi, Inversions in Compositions of Integers, to appear in Quaestiones Mathematicae.

Row sums are A189052. The first column is A004526(n-1). Diagonal is A000217(n-2). Lower diagonal is A002411(n-3). 2nd lower diagonal is A001621(n-4).

T:= proc(n, k) option remember;
      if k=2 then floor((n-1)/2)
    elif k>=n then 0
    else T(n-1, k) +k/(k-2) *T(n-1, k-1)
      fi
    end:
seq(seq(T(n, k), k=2..n-1), n=3..13);  # Alois P. Heinz, Apr 17 2011

T[n_, k_] := T[n, k] = Which[k == 2, Floor[(n-1)/2], k >= n, 0, True, T[n-1, k] + k/(k-2)*T[n-1, k-1]]; Table[Table[T[n, k], {k, 2, n-1}], {n, 3, 13}] // Flatten (* Jean-François Alcover, Jan 14 2014, after Alois P. Heinz *)

G.f.: (1-x)*x^3/((1+x)*(1-x-y*x)^3). - Geoffrey Critzer, Mar 19 2014

A277792 Squares that are also pentagonal pyramidal numbers.

Original entry on oeis.org

0, 1, 196, 2601, 15376, 60025, 181476, 461041, 1032256, 2099601, 3960100, 7027801, 11861136, 19193161, 29964676, 45360225, 66846976, 96216481, 135629316, 187662601, 255360400, 342287001, 452583076, 591024721, 763085376, 975000625, 1233835876, 1547556921, 1925103376, 2376465001, 2912760900

Offset: 0

Ilya Gutkovskiy, Oct 31 2016

Intersection of A000290 and A002411.

			a(2) = 196 because 196 = 14^2 is a perfect square and 196 = 7^2*(7 + 1)/2 is the 7th pentagonal pyramidal number.

Daniel Mondot, Table of n, a(n) for n = 0..1000
Eric Weisstein's World of Mathematics, Pentagonal Pyramidal Number
Eric Weisstein's World of Mathematics, Square Number
Index to sequences related to polygonal numbers
Index to sequences related to pyramidal numbers
Index entries for two-way infinite sequences
Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,1).

Cf. A000290, A001653, A002411, A007588, A036353, A056220, A254333.

[n^2*(2*n^2-1)^2: n in [0..30]]; // Vincenzo Librandi, Nov 01 2016

Table[n^2 (2 n^2 - 1)^2, {n, 0, 30}]
LinearRecurrence[{7,-21,35,-35,21,-7,1},{0,1,196,2601,15376,60025,181476},40] (* Harvey P. Dale, Nov 01 2024 *)

O.g.f.: x*(1 + 189*x + 1250*x^2 + 1250*x^3 + 189*x^4 + x^5)/(1 - x)^7.
E.g.f.: x*(1 + 97*x + 336*x^2 + 256*x^3 + 60*x^4 + 4*x^5)*exp(x).
a(n) = a(-n).
a(n) = n^2*(2*n^2 - 1)^2.
a(n) = A000290(A007588(n)).
a(n) = A000290(n)*A000290(A056220(n)).
Sum_{n>=1} 1/a(n) = (2*Pi^2+9*sqrt(2)*Pi*cot(Pi/sqrt(2))+3*Pi^2*csc(Pi/sqrt(2))^2-24)/12 = 1.0055779712856...

A299412 Pentagonal pyramidal numbers divisible by 3.

Original entry on oeis.org

0, 6, 18, 75, 126, 288, 405, 726, 936, 1470, 1800, 2601, 3078, 4200, 4851, 6348, 7200, 9126, 10206, 12615, 13950, 16896, 18513, 22050, 23976, 28158, 30420, 35301, 37926, 43560, 46575, 53016, 56448, 63750, 67626, 75843, 80190, 89376, 94221, 104430, 109800, 121086, 127008, 139425, 145926, 159528, 166635

Offset: 0

Justin Gaetano, Feb 20 2018

nonn

			The first 6 pentagonal pyramidal numbers are 0, 1, 6, 18, 40, 75; of these, 0, 6, 18, 75 are divisible by 3.

Justin Gaetano, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (1,3,-3,-3,3,1,-1)

Cf. A001318, A002411, A007494, A117748, A269416.

[IsEven(n) select (3*n/2)^2*(3*n/2+1)/2 else ((3*n+1)/2)^2*((3*n+1)/2+1)/2: n in [0..50] ]; // Vincenzo Librandi, Mar 14 2018

f:= proc(n) if n::even then (3*n/2)^2*(3*n/2+1)/2 else
((3*n+1)/2)^2*((3*n+1)/2+1)/2 fi end proc:
map(f, [$0..100]); # Robert Israel, Feb 28 2018

Array[((3 #1 + #2)/2)^2*((3 #1 + #2)/2 + 1)/2 & @@ {#, Boole@ OddQ@ #} &, 47, 0] (* Michael De Vlieger, Feb 21 2018 *)
LinearRecurrence[{1,3,-3,-3,3,1,-1},{0,6,18,75,126,288,405},50] (* Harvey P. Dale, Jul 16 2021 *)

lista(nn) = {for (n=0, nn, if (!(n^2*(n+1)/2 % 3), print1(n^2*(n+1)/2, ", ")););} \\ Michel Marcus, Feb 21 2018

x='x+O('x^99); concat(0, Vec(3*x*(3*x^4+5*x^3+13*x^2+4*x+2)/((x-1)^4*(x+1)^3))) \\ Altug Alkan, Mar 14 2018

a(n) = A007494(n)*A117748(n).
a(n) = (3*n/2)^2*(3*n/2+1)/2 if n even.
a(n) = ((3*n+1)/2)^2*((3*n+1)/2+1)/2 if n odd.
From Omar E. Pol, Feb 21 2018: (Start)
a(n) = 3*A001318(n)*A007494(n).
a(n) = A001318(n)*abs(A269416(n-1)), n >= 1. (End)
G.f.: 3*x*(3*x^4 + 5*x^3 + 13*x^2 + 4*x + 2)/((x-1)^4*(x+1)^3). - Robert Israel, Feb 28 2018

A309176 a(n) = n^2 * (n + 1)/2 - Sum_{k=1..n} sigma_2(k).

Original entry on oeis.org

0, 0, 2, 3, 12, 13, 33, 40, 66, 81, 135, 135, 212, 249, 319, 354, 489, 511, 681, 725, 876, 981, 1233, 1235, 1509, 1660, 1920, 2032, 2437, 2472, 2936, 3091, 3488, 3755, 4275, 4290, 4955, 5292, 5854, 6024, 6843, 6968, 7870, 8190, 8839, 9340, 10420, 10442, 11568, 12038, 13014, 13474, 14851, 15098, 16436

Offset: 1

Ilya Gutkovskiy, Jul 15 2019

nonn

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A000326, A001157, A004125, A048158, A051126, A154585, A256532.

Cf. A002411, A064602.

Table[n^2 (n + 1)/2 - Sum[DivisorSigma[2, k], {k, 1, n}], {n, 1, 55}]
nmax = 55; CoefficientList[Series[x (1 + 2 x)/(1 - x)^4 - 1/(1 - x) Sum[k^2 x^k/(1 - x^k), {k, 1, nmax}], {x, 0, nmax}], x] // Rest
Table[Sum[Mod[n, k] k, {k, 1, n}], {n, 1, 55}]

a(n) = n^2*(n+1)/2 - sum(k=1, n, sigma(k, 2)); \\ Michel Marcus, Sep 18 2021

from math import isqrt
def A309176(n): return (n**2*(n+1)>>1)+((s:=isqrt(n))**2*(s+1)*(2*s+1)-sum((q:=n//k)*(6*k**2+q*(2*q+3)+1) for k in range(1,s+1)))//6 # Chai Wah Wu, Oct 21 2023

G.f.: x * (1 + 2*x)/(1 - x)^4 - (1/(1 - x)) * Sum_{k>=1} k^2 * x^k/(1 - x^k).
a(n) = Sum_{k=1..n} (n mod k) * k.
a(n) = A002411(n) - A064602(n).

A343558 Irregular triangle read by rows: the n-th row gives the row indices of the consecutive elements of the spiral of the n X n matrix defined in A126224.

Original entry on oeis.org

1, 1, 1, 2, 2, 1, 1, 1, 2, 3, 3, 3, 2, 2, 1, 1, 1, 1, 2, 3, 4, 4, 4, 4, 3, 2, 2, 2, 3, 3, 1, 1, 1, 1, 1, 2, 3, 4, 5, 5, 5, 5, 5, 4, 3, 2, 2, 2, 2, 3, 4, 4, 4, 3, 3, 1, 1, 1, 1, 1, 1, 2, 3, 4, 5, 6, 6, 6, 6, 6, 6, 5, 4, 3, 2, 2, 2, 2, 2, 3, 4, 5, 5, 5, 5, 4, 3, 3, 3, 4, 4

Offset: 1

Stefano Spezia, Apr 19 2021

			The triangle begins
1
1   1   2   2
1   1   1   2   3   3   3   2   2
1   1   1   1   2   3   4   4   4   4   3   2   2   2   3   3
...

Stefano Spezia, First 30 rows of the triangle, flattened

Cf. A000290 (row length), A002265, A002411 (row sums), A010873, A060747, A126224, A343559 (column indices).

a:={};nmax:=6;For[n=1,n<=nmax,n++,For[s=1,s<=2n-1,s++,If[OddQ[s] &&Mod[s,4]==1,k=Ceiling[s/4];For[i=1,i<=Ceiling[n-s/2],i++,AppendTo[a,k]],If[EvenQ[s]&&Mod[s,4]==2,For[i=1,i<=Ceiling[n-s/2],i++,AppendTo[a,k+i]];k+=Ceiling[n-s/2],If[EvenQ[s]&&Mod[s,4]==0,For[i=1,i<=Ceiling[n-s/2],i++,AppendTo[a,k-i]];k=k-i+1,For[i=1,i<=Ceiling[n-s/2],i++,AppendTo[a,k]]]]]]];a

A343559 Irregular triangle read by rows: the n-th row gives the column indices of the consecutive elements of the spiral of the n X n matrix defined in A126224.

Original entry on oeis.org

1, 1, 2, 2, 1, 1, 2, 3, 3, 3, 2, 1, 1, 2, 1, 2, 3, 4, 4, 4, 4, 3, 2, 1, 1, 1, 2, 3, 3, 2, 1, 2, 3, 4, 5, 5, 5, 5, 5, 4, 3, 2, 1, 1, 1, 1, 2, 3, 4, 4, 4, 3, 2, 2, 3, 1, 2, 3, 4, 5, 6, 6, 6, 6, 6, 6, 5, 4, 3, 2, 1, 1, 1, 1, 1, 2, 3, 4, 5, 5, 5, 5, 4, 3, 2, 2, 2, 3, 4, 4, 3

Offset: 1

Stefano Spezia, Apr 19 2021

			The triangle begins
1
1   2   2   1
1   2   3   3   3   2   1   1   2
1   2   3   4   4   4   4   3   2   1   1   1   2   3   3   2
...

Stefano Spezia, First 30 rows of the triangle, flattened

Cf. A000290 (row length), A002265, A002411 (row sums), A010873, A060747, A126224, A343558 (row indices).

a:={};nmax:=6;For[n=1,n<=nmax,n++,For[s=1,s<=2n-1,s++,If[OddQ[s]&&Mod[s,4]==1 ,k=Floor[s/4];For[i=1,i<=Ceiling[n-s/2],i++,AppendTo[a,k+i]];k=+Floor[s/4]+Ceiling[n-s/2],If[EvenQ[s],For[i=1,i<=Ceiling[n-s/2],i++,AppendTo[a,k]],For[i=1,i<=Ceiling[n-s/2],i++,AppendTo[a,k-i]];k=k-i+1]]]]; a

A343853 Irregular triangle read by rows: the n-th row gives the row indices of the matrix of 1..n^2 filled successively back and forth along antidiagonals.

Original entry on oeis.org

1, 1, 1, 2, 2, 1, 1, 2, 3, 2, 1, 2, 3, 3, 1, 1, 2, 3, 2, 1, 1, 2, 3, 4, 4, 3, 2, 3, 4, 4, 1, 1, 2, 3, 2, 1, 1, 2, 3, 4, 5, 4, 3, 2, 1, 2, 3, 4, 5, 5, 4, 3, 4, 5, 5, 1, 1, 2, 3, 2, 1, 1, 2, 3, 4, 5, 4, 3, 2, 1, 1, 2, 3, 4, 5, 6, 6, 5, 4, 3, 2, 3, 4, 5, 6, 6, 5, 4, 5, 6, 6

Offset: 1

Stefano Spezia, May 01 2021

			The triangle begins:
1
1   1   2   2
1   1   2   3   2   1   2   3   3
1   1   2   3   2   1   1   2   3   4   4   3   2   3   4   4
...

Stefano Spezia, First 30 rows of the triangle, flattened

Cf. A000290 (row length), A002411 (row sums), A060747 (number of antidiagonals), A078475, A319573, A343854 (column indices).

a={};For[n=1,n<=6,n++,For[d=1,d<=n,d++, If[OddQ[d],i=d;For [k=1,k<=d,k++, AppendTo[a,i-k+1]],i=1;For[k=1,k<=d,k++, AppendTo[a,i+k-1]]]];For[d=n+1,d<=2n-1,d++, If[OddQ[d],i= n; For[k=1,k<=2n-d,k++,AppendTo[a,i-k+1]],If[EvenQ[d],i=d-n+1;For[k=1,k<=2n-d,k++, AppendTo[a,i+k-1]]]]]]; a

A343854 Irregular triangle read by rows: the n-th row gives the column indices of the matrix of 1..n^2 filled successively back and forth along antidiagonals.

Original entry on oeis.org

1, 1, 2, 1, 2, 1, 2, 1, 1, 2, 3, 3, 2, 3, 1, 2, 1, 1, 2, 3, 4, 3, 2, 1, 2, 3, 4, 4, 3, 4, 1, 2, 1, 1, 2, 3, 4, 3, 2, 1, 1, 2, 3, 4, 5, 5, 4, 3, 2, 3, 4, 5, 5, 4, 5, 1, 2, 1, 1, 2, 3, 4, 3, 2, 1, 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1, 2, 3, 4, 5, 6, 6, 5, 4, 3, 4, 5, 6, 6, 5, 6

Offset: 1

Stefano Spezia, May 01 2021

			The triangle begins:
1
1   2   1   2
1   2   1   1   2   3   3   2   3
1   2   1   1   2   3   4   3   2   1   2   3   4   4   3   4
...

Stefano Spezia, First 30 rows of the triangle, flattened

Cf. A000290 (row length), A002411 (row sums), A060747 (number of antidiagonals), A078475, A319572, A343853 (row indices).

a={};For[n=1,n<=6,n++,For[d=1,d<=n,d++, If[EvenQ[d],i=d;For [k=1,k<=d,k++, AppendTo[a,i-k+1]],i=1;For[k=1,k<=d,k++, AppendTo[a,i+k-1]]]];For[d=n+1,d<=2n-1,d++, If[EvenQ[d],i= n; For[k=1,k<=2n-d,k++,AppendTo[a,i-k+1]],If[OddQ[d],i=d-n+1;For[k=1,k<=2n-d,k++, AppendTo[a,i+k-1]]]]]]; a

cp's OEIS Frontend

A136360 Square roots of the perfect squares in A133459.

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A136361 Square roots of the perfect squares in A136360; or numbers k such that k^4 is in A133459 = the sums of two nonzero pentagonal pyramidal numbers.

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A189073 Triangle read by rows: T(n,k) is the number of inversions in k-compositions of n for n >= 3, 2 <= k <= n-1.

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Maple

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A277792 Squares that are also pentagonal pyramidal numbers.

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Magma

Mathematica

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A299412 Pentagonal pyramidal numbers divisible by 3.

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Magma

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PARI

PARI

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A309176 a(n) = n^2 * (n + 1)/2 - Sum_{k=1..n} sigma_2(k).

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A343558 Irregular triangle read by rows: the n-th row gives the row indices of the consecutive elements of the spiral of the n X n matrix defined in A126224.

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A343559 Irregular triangle read by rows: the n-th row gives the column indices of the consecutive elements of the spiral of the n X n matrix defined in A126224.

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