Daniel Poveda Parrilla - cp's OEIS frontend

A288487 Cuboids that fit in square rings from A288486 obtaining a fifth power.

1, 8, 75, 400, 1445, 4056, 9583, 20000, 38025, 67240, 112211, 178608, 273325, 404600, 582135, 817216, 1122833, 1513800, 2006875, 2620880, 3376821, 4298008, 5410175, 6741600, 8323225, 10188776, 12374883, 14921200, 17870525, 21268920, 25165831, 29614208

Offset: 0

Daniel Poveda Parrilla, Jun 11 2017

If we add a(n) and A288487(n) graphically we obtain a bigger cuboid which is a square of cubes (a cuboid with dimensions n^2 * n^2 * n).

a(10^n) is a palindrome in base 10.

Daniel Poveda Parrilla, Table of n, a(n) for n = 0..10000
Daniel Poveda Parrilla, Illustration of initial terms
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Cf. A000584, A002061, A002522, A014206, A059100, A082044, A288486.

Table[(1 + n)*(1 + n^2)^2, {n, 0, 28}] (* or *) CoefficientList[Series[(1 + 2 x + 42 x^2 + 50 x^3 + 25 x^4)/(1 - x)^6, {x, 0, 28}], x] (* or *) LinearRecurrence[{6, -15, 20, -15, 6, -1}, {1, 8, 75, 400, 1445, 4056}, 29]

Vec((1 + 2*x + 42*x^2 + 50*x^3 + 25*x^4)/(1 - x)^6 + O(x^28))

G.f.: (1 + 2*x + 42*x^2 + 50*x^3 + 25*x^4)/(1 - x)^6.
a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6).
a(n) = (n + 1)*(n^2 + 1)^2 = (n + 1)*(A002522(n))^2 = (n + 1)*A082044(n).
a(n) = n^5 + A002061(A002061(n+1)).
a(n) = A000584(n+1) - A288486(n)
a(n) = (n + 1)*A059100(n-1) + 4*(n^2 -1)*A014206(n-1) for n > 1.

A288486 Square rings obtained by adding four identical cuboids from A169938, a(n) = 4n(n+1)(n(n+1)+1).

Original entry on oeis.org

0, 24, 168, 624, 1680, 3720, 7224, 12768, 21024, 32760, 48840, 70224, 97968, 133224, 177240, 231360, 297024, 375768, 469224, 579120, 707280, 855624, 1026168, 1221024, 1442400, 1692600, 1974024, 2289168, 2640624, 3031080, 3463320, 3940224, 4464768, 5040024

Offset: 0

Daniel Poveda Parrilla, Jun 10 2017

If we fill the empty space with A288487(n) cubes, we get a solid cuboid with (n+1)^5 cubes (A000584(n+1)).

Daniel Poveda Parrilla, Table of n, a(n) for n = 0..10000
Daniel Poveda Parrilla, Illustration of initial terms
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A000584, A001844, A002061, A002378, A007204, A033996, A169938, A288487.

Table[4 n (n + 1) (n^2 + n + 1), {n, 0, 28}] (* or *) CoefficientList[Series[24 (x + 2 x^2 + x^3)/(1 - x)^5, {x, 0, 28}], x] (* or *) LinearRecurrence[{5, -10, 10, -5, 1}, {0, 24, 168, 624, 1680}, 29]

Vec(24*(x + 2*x^2 + x^3)/(1 - x)^5 + O(x^28))

G.f.: 24*(x + 2*x^2 + x^3) / (1 - x)^5.
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).
a(n) = A007204(n) - 1 = (A001844(n))^2 - 1.
a(n) = 4*A169938(n) = 4*A002378(n)*A002061(n+1) = A033996(n)*A002061(n+1).

A282390 Width of polyominoes in A282389.

Original entry on oeis.org

3, 5, 8, 14, 27, 53, 104, 206, 410, 818, 1635, 3269, 6536, 13070, 26139, 52277, 104552, 209102, 418202, 836402, 1672803, 3345605, 6691209, 13382417, 26764832, 53529662, 107059322, 214118642, 428237283, 856474565, 1712949128, 3425898254, 6851796507

Offset: 1

Daniel Poveda Parrilla, Feb 14 2017

nonn

Polyominoes in A282389 have got a width of a(n+1) squares and a height of A000051(n) squares.

The polyomino may be represented as a sequence of the lengths of steps in the "ladder" of the polyomino: [2, 1] for L-tetromino, [2, 1, 2] for the next iteration, and so on. The overall width is the sum of these lengths. And on the next iteration, the new sequence of lengths of steps is formed from the previous one as: + . So the sequence always consists of 1's and 2's only and therefore can be encoded as a binary string of length 2^n+1. This is exploited in the Python program below and explains the formula. - Andrey Zabolotskiy, Feb 14 2017

			a(1) = 3
a(2) = 2 * 3 - 1 = 5
a(3) = 2 * 5 - 2 = 8
a(4) = 2 * 8 - 2 = 14
a(5) = 2 * 14 - 1 = 27
a(6) = 2 * 27 - 1 = 53
a(7) = 2 * 53 - 2 = 104
a(8) = 2 * 104 - 2 = 206
a(9) = 2 * 206 - 2 = 410
a(10) = 2 * 410 - 2 = 818
a(11) = 2 * 818 - 1 = 1635
a(12) = 2 * 1635 - 1 = 3269
a(13) = 2 * 3269 - 2 = 6536
a(14) = 2 * 6536 - 2 = 13070
a(15) = 2 * 13070 - 1 = 26139
a(16) = 2 * 26139 - 1 = 52277
a(17) = 2 * 52277 - 2 = 104552
a(18) = 2 * 104552 - 2 = 209102

Andrey Zabolotskiy, Table of n, a(n) for n = 1..3300

Cf. A282389.

w, h, bp, bp2 = 3, 2, 0b10, 0b01
for i in range(1, 10):
    print(w)
    w, h, bp, bp2 = w*2-(2 if (bp&1) else 1), 2**i+1, ((bp2&((1<<(h-1))-1))<>1)
for i in range(100):
    print(w)
    w, h, bp, bp2 = w*2-(2 if (bp&1) else 1), h-1, bp2, (bp>>1)
# Andrey Zabolotskiy, Feb 14 2017

a(1) = 3, a(n) = 2*a(n-1) - k for n > 1, where k is the width of the central step in the "ladder", which is 1 or 2.

A282389 Number of squares in triangle-shaped polyominoes obtained by adding three identical polyominoes to the previous one, starting with one L-tetromino.

Original entry on oeis.org

4, 10, 25, 70, 238, 901, 3445, 13390, 52942, 210226, 838450, 3350725, 13393093, 53547790, 214151950, 856558645, 3426077749, 13703917774, 54815043790, 219258602290, 877031899954, 3508124454085, 14032487779525, 56129938535185, 224519713993489, 898078755310654

Offset: 0

Daniel Poveda Parrilla, Feb 14 2017

nonn

Each polyomino a(n) has a width of A282390(n+1) squares and a height of A000051(n) squares.

Daniel Poveda Parrilla, Illustration of initial terms

Cf. A000051, A028401, A282390.

m = [3, 1]
for i in range(25):
    w, h = m[0], len(m)
    print(sum(m)) # print(w) for widths of the polyominoes
    m2 = [w-x for x in reversed(m) if w>x]
    m = [w+x for x in m2] + [w for x in m if x==w] + m2
# Andrey Zabolotskiy, Feb 14 2017

a(0) = 4; a(n) = a(n-1) + 3*(A282390(n)*A000051(n-1) - a(n-1)) for n > 0.
a(n) = A282390(n+(n mod 2))*A000051(n-(n mod 2)) for n > 0.
a(n) = (A282390(n+(n mod 2)) - A000051(n-1+(n mod 2)))*A000051(n-(n mod 2)) + A028401(n+2) for n > 0.

A276918 a(2n) = A060867(n+1), a(2n+1) = A092440(n+1).

Original entry on oeis.org

1, 5, 9, 25, 49, 113, 225, 481, 961, 1985, 3969, 8065, 16129, 32513, 65025, 130561, 261121, 523265, 1046529, 2095105, 4190209, 8384513, 16769025, 33546241, 67092481, 134201345, 268402689, 536838145, 1073676289, 2147418113, 4294836225, 8589803521, 17179607041

Offset: 0

Daniel Poveda Parrilla, Jan 26 2017

nonn

In binary there is a pattern in how the zeros and ones appear:
a(0)  =         01
a(1)  =        101
a(2)  =        1001
a(3)  =       11001
a(4)  =       110001
a(5)  =      1110001
a(6)  =      11100001
a(7)  =     111100001
a(8)  =     1111000001
a(9)  =    11111000001
a(10) =    111110000001
a(11) =   1111110000001
a(12) =   11111100000001
a(13) =  111111100000001
a(14) =  1111111000000001
a(15) = 11111111000000001
Graphically, each term can be obtained by successively and alternately forming squares and centered squares as shown in the illustration.

Daniel Poveda Parrilla, Table of n, a(n) for n = 0..1000
Daniel Poveda Parrilla, Illustration of initial terms
Index entries for linear recurrences with constant coefficients, signature (3,0,-6,4).

Cf. A000225, A060867, A092440, A224195.

Table[1+2^(n+2)-2^(1+n/2)+(-1)^(n+1) 2^(1+n/2)-2^((n+1)/2)+(-1)^(n+2) 2^((n+1)/2), {n,0,28}] (*or*)
CoefficientList[Series[(-1 - 2 x + 6 x^2 - 4 x^3)/(-1 + 3 x - 6 x^3 + 4 x^4), {x,0,28}], x] (*or*)
LinearRecurrence[{3, 0, -6, 4}, {1, 5, 9, 25}, 29]

Vec((-1-2*x+6*x^2-4*x^3) / (-1+3*x-6*x^3+4*x^4) + O(x^29))

a(n) = 1 + 2^(n+2) - 2^(1 + n/2) + (-1)^(n+1)*2^(1 + n/2) - 2^((n+1)/2) + (-1)^(n+2)*2^((n+1)/2).
a(n) = 3*a(n-1) - 6*a(n-3) + 4*a(n-4) for n>3.
G.f.: (-1-2*x+6*x^2-4*x^3)/(-1+3*x-6*x^3+4*x^4).

A276917 Numbers obtained by alternatively adding centered pentagonal layers of 5(2^n-1) and 5(3^n-1) elements.

Original entry on oeis.org

1, 6, 16, 31, 71, 106, 236, 311, 711, 866, 2076, 2391, 6031, 6666, 17596, 18871, 51671, 54226, 152636, 157751, 452991, 463226, 1348956, 1369431, 4026631, 4067586, 12039196, 12121111, 36035951, 36199786, 107944316, 108271991, 323505591, 324160946, 969861756

Offset: 0

Daniel Poveda Parrilla, Dec 29 2016

a(0), a(1), a(2) and a(3) are the first four centered pentagonal numbers, as they match the same pattern. From a(4) onwards all terms are a different kind of centered pentagonal numbers, as the number of elements in subsequent layers doesn't increase uniformly.
a(13) is the first palindromic number in the sequence. a(19) is the second one.
First prime terms are a(3), a(4), a(7), a(31), a(100) and a(115).

Daniel Poveda Parrilla, Table of n, a(n) for n = 0..1000
Daniel Poveda Parrilla, Illustration of initial terms
Index entries for linear recurrences with constant coefficients, signature (2,4,-10,-1,12,-6).

Cf. A005891.

Table[5 (Sum[2^i, {i, 0, ((n + Mod[n, 2])/2)}] + Sum[3^j, {j, 0, ((n - Mod[n, 2])/2)}]) - 5 n - 9, {n, 0, 28}] (* or *)
CoefficientList[Series[(1 + 4 x - 15 x^3 + 6 x^4 - 6 x^5)/((-1 + x)^2 (1 - 5 x^2 + 6 x^4)), {x, 0, 28}], x] (* or *)
LinearRecurrence[{2, 4, -10, -1, 12, -6}, {1, 6, 16, 31, 71, 106}, 29]

Vec((1+4*x-15*x^3+6*x^4-6*x^5) / ((-1+x)^2*(1-5*x^2+6*x^4)) + O(x^40)) \\ Colin Barker, Dec 30 2016

a(n) = 5*(Sum_{i=0..((n+(n mod 2))/2)} 2^i + Sum_{j=0..((n-(n mod 2))/2)} 3^j) - 5*n - 9.
a(n) = a(n-1) + 5*((2+((n+1) mod 2))^((n+(n mod 2))/2) - 1) for n>0.
G.f.: (1+4*x-15*x^3+6*x^4-6*x^5)/((-1+x)^2*(1-5*x^2+6*x^4)).
From Colin Barker, Dec 30 2016: (Start)
a(n) = (-10*n + 5*3^(n/2+1) + 5*2^(n/2+2) - 33)/2 for n even.
a(n) = (-10*n + 5*3^(n/2+1/2) + 5*2^(n/2+5/2) - 33)/2 for n odd.
(End)

A276915 Indices of triangular numbers in A276914 which are also pentagonal.

Original entry on oeis.org

0, 1, 10, 143, 1988, 27693, 385710, 5372251, 74825800, 1042188953, 14515819538, 202179284583, 2815994164620, 39221739020101, 546288352116790, 7608815190614963, 105977124316492688, 1476070925240282673, 20559015829047464730, 286350150681424223551

Offset: 0

Daniel Poveda Parrilla, Sep 22 2016

A276914(a(n)) = A014979(n + 1). All numbers which are both triangular and pentagonal can be found in sequence A276914.

Daniel Poveda Parrilla, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (13,13,-1).

Cf. A014979, A046175, A276914.

RecurrenceTable[{a[n] == 14 a[n - 1] - a[n - 2] - 4 (-1)^n, a[0] == 0, a[1] == 1}, a, {n, 19}] (* Michael De Vlieger, Sep 23 2016 *)

concat(0, Vec(x*(1-3*x)/((1+x)*(1-14*x+x^2)) + O(x^30))) \\ Colin Barker, Sep 23 2016

a(n) = 14*a(n-1) - a(n-2) - 4*(-1)^n for n>1, a(0)=0, a(1)=1.
a(n) = (A046175(n) + (A046175(n) mod 2))/2.
From Colin Barker, Sep 23 2016: (Start)
G.f.: x*(1 - 3*x) / ((1 + x)*(1 - 14*x + x^2)).
a(n) = 13*a(n-1) + 13*a(n-2) - a(n-3) for n>2.
a(n) = ( -6*(-1)^n + (3+sqrt(3))*(7-4*sqrt(3))^n - (-3+sqrt(3))*(7+4*sqrt(3))^n )/24. (End)

A276916 Subsequence of centered square numbers obtained by adding four triangles from A276914 and a central element, a(n) = 4*A276914(n) + 1.

Original entry on oeis.org

1, 5, 41, 61, 145, 181, 313, 365, 545, 613, 841, 925, 1201, 1301, 1625, 1741, 2113, 2245, 2665, 2813, 3281, 3445, 3961, 4141, 4705, 4901, 5513, 5725, 6385, 6613, 7321, 7565, 8321, 8581, 9385, 9661, 10513, 10805, 11705, 12013, 12961, 13285, 14281, 14621, 15665

Offset: 0

Daniel Poveda Parrilla, Sep 27 2016

All terms of this sequence are centered square numbers. Graphically, each term of the sequence is made of four squares, eight triangles and a central element.

a(A220185(n+1)) = A008844(2n) = A079291(4n+1), which is a square of a Pell number.

Daniel Poveda Parrilla, Table of n, a(n) for n = 0..10000
Daniel Poveda Parrilla, Illustration of initial terms
Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Cf. A008844, A079291, A220185, A276914.

[4*n*(2*n+(-1)^n)+1 : n in [0..60]]; // Wesley Ivan Hurt, Sep 27 2016

A276916:=n->4*n*(2*n+(-1)^n)+1: seq(A276916(n), n=0..60); # Wesley Ivan Hurt, Sep 27 2016

Table[4 n (2 n + (-1)^n) + 1, {n, 0, 44}] (* or *)
CoefficientList[Series[(1 +4x +34x^2 +12x^3 +13x^4)/((1-x)^3*(1+x)^2), {x, 0, 44}], x] (* Michael De Vlieger, Sep 28 2016 *)

Vec((1+4*x+34*x^2+12*x^3+13*x^4)/((1-x)^3*(1+x)^2) + O(x^50)) \\ Colin Barker, Sep 27 2016

[4*n*(2*n+(-1)^n) +1 for n in (0..60)] # G. C. Greubel, Aug 19 2022

a(n) = 4*n*(2*n + (-1)^n) + 1.
a(n) = 4*n*(2*n + 1) + 1 for n even.
a(n) = 4*n*(2*n - 1) + 1 for n odd.
a(n) is sum of two squares; a(n) = k^2 + (k+1)^2 where k = 2n-(n mod 2). - David A. Corneth, Sep 27 2016
From Colin Barker, Sep 27 2016: (Start)
a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5) for n > 4.
G.f.: (1+4*x+34*x^2+12*x^3+13*x^4) / ((1-x)^3*(1+x)^2). (End)
E.g.f.: (1+8*x+8*x^2)*exp(x) - 4*x*exp(-x). - G. C. Greubel, Aug 19 2022

A276914 Subsequence of triangular numbers obtained by adding a square and two smaller triangles, a(n) = n^2 + 2*A000217(A052928(n)).

Original entry on oeis.org

0, 1, 10, 15, 36, 45, 78, 91, 136, 153, 210, 231, 300, 325, 406, 435, 528, 561, 666, 703, 820, 861, 990, 1035, 1176, 1225, 1378, 1431, 1596, 1653, 1830, 1891, 2080, 2145, 2346, 2415, 2628, 2701, 2926, 3003, 3240, 3321, 3570, 3655, 3916, 4005, 4278, 4371, 4656

Offset: 0

Daniel Poveda Parrilla, Sep 22 2016

All terms of this sequence are triangular numbers. Graphically, for each term of the sequence, one corner of the square will be part of the corresponding triangle's hypotenuse if the term is an odd number. Otherwise, it will not be part of it.
a(A276915(n)) is a triangular pentagonal number.
a(A079291(n)) is a triangular square number, as A275496 is a subsequence of this.

Daniel Poveda Parrilla, Table of n, a(n) for n = 0..10000
Daniel Poveda Parrilla, Illustration of initial terms.
Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Cf. A000217, A004526, A016813, A042948, A052928, A079291, A168277, A275496, A276915.

[n*(2*n+(-1)^n): n in [0..40]]; // G. C. Greubel, Aug 19 2022

Table[n (2 n + (-1)^n), {n, 0, 48}] (* Michael De Vlieger, Sep 23 2016 *)

concat(0, Vec(x*(1+9*x+3*x^2+3*x^3)/((1-x)^3*(1+x)^2) + O(x^50))) \\ Colin Barker, Sep 23 2016

[n*(2*n+(-1)^n) for n in (0..40)] # G. C. Greubel, Aug 19 2022

a(n) = n^2 + 2*A000217(A052928(n)).
a(n) = A000217(A042948(n)).
a(n) = n*(2*n + (-1)^n).
a(n) = n*A168277(n + 1).
a(n) = n*A016813(A004526(n)).
From Colin Barker, Sep 23 2016: (Start)
G.f.: x*(1 + 9*x + 3*x^2 + 3*x^3) / ((1 - x)^3*(1 + x)^2).
a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5) for n>4.
a(n) = n*(2*n+1) for n even.
a(n) = n*(2*n-1) for n odd. (End)
E.g.f.: x*( 2*(1+x)*exp(x) - exp(-x) ). - G. C. Greubel, Aug 19 2022
Sum_{n>=1} 1/a(n) = 2 - log(2). - Amiram Eldar, Aug 21 2022

A275543 A081585 and A069129 interleaved.

Original entry on oeis.org

1, 1, 9, 17, 33, 49, 73, 97, 129, 161, 201, 241, 289, 337, 393, 449, 513, 577, 649, 721, 801, 881, 969, 1057, 1153, 1249, 1353, 1457, 1569, 1681, 1801, 1921, 2049, 2177, 2313, 2449, 2593, 2737, 2889, 3041, 3201, 3361, 3529, 3697, 3873, 4049, 4233, 4417, 4609

Offset: 0

Daniel Poveda Parrilla, Aug 01 2016

a(A000129(n)) is a square.
(n^2)*a(n) = A275496(n) which is a triangular number.
(A000129(n)^2)*a(A000129(n)) = A275496(A000129(n)) = A001110(n) which is a square triangular number.
a(2n+1)/a(2n) is convergent to 1.

			a(1) = A275496(1) = 1.
a(5) = A275496(5)/25 = 1225/25 = 49.
a(7) = A275496(7)/49 = 4753/49 = 97.
a(12) = A275496(12)/144 = 41616/144 = 289.

Daniel Poveda Parrilla, Table of n, a(n) for n = 0..500000
Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).

Cf. A081585(n) = a(2n), A069129(n) = a(2n + 1).

Cf. A000129, A000217, A000290, A001110, A077221, A093954, A275496.

CoefficientList[Series[(1 - x + 7 x^2 + x^3)/((1 - x)^3 (1 + x)), {x, 0, 48}], x] (* or as defined *)
Riffle[LinearRecurrence[{3, -3, 1}, {1, 9, 33}, #], FoldList[#1 + #2 &, 1, 16 Range@ #]] &@ 25 (* Michael De Vlieger, Aug 01 2016, after Vincenzo Librandi at A081585 and Robert G. Wilson v at A069129 *)

a(n)=(-1)^n + 2*n^2 \\ Charles R Greathouse IV, Aug 03 2016

Vec((1-x+7*x^2+x^3)/((1-x)^3*(1+x)) + O(x^100)) \\ Colin Barker, Aug 21 2016

a(0) = 1; a(n) = A275496(n)/(n^2) for n > 0.
From Colin Barker, Aug 01 2016: (Start)
a(n) = (2*n^2 + (-1)^n).
a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4) for n > 3.
G.f.: (1 -x +7*x^2 +x^3) / ((1 - x)^3*(1 + x)).
(End)
From Daniel Poveda Parrilla, Aug 18 2016: (Start)
a(2n) = A077221(2n) + 1.
a(2n + 1) = A077221(2n + 1). (End)
Sum_{n>=0} 1/a(n) = (1 + (tan(c) + coth(c))*c)/2, where c = Pi/(2*sqrt(2)) is A093954. - Amiram Eldar, Aug 21 2022

cp's OEIS Frontend

User: Daniel Poveda Parrilla

A288487 Cuboids that fit in square rings from A288486 obtaining a fifth power.

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A288486 Square rings obtained by adding four identical cuboids from A169938, a(n) = 4*n*(n+1)*(n*(n+1)+1).

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A282390 Width of polyominoes in A282389.

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A282389 Number of squares in triangle-shaped polyominoes obtained by adding three identical polyominoes to the previous one, starting with one L-tetromino.

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A276918 a(2n) = A060867(n+1), a(2n+1) = A092440(n+1).

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A276917 Numbers obtained by alternatively adding centered pentagonal layers of 5*(2^n-1) and 5*(3^n-1) elements.

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A276915 Indices of triangular numbers in A276914 which are also pentagonal.

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Formula

A276916 Subsequence of centered square numbers obtained by adding four triangles from A276914 and a central element, a(n) = 4*A276914(n) + 1.

Views

A288486 Square rings obtained by adding four identical cuboids from A169938, a(n) = 4n(n+1)(n(n+1)+1).

A276917 Numbers obtained by alternatively adding centered pentagonal layers of 5(2^n-1) and 5(3^n-1) elements.