A262710 -id:A262710 - cp's OEIS frontend

A359452 Number of vertices in the partite set of the n-Menger sponge graph that contains the corners.

1, 8, 208, 3968, 80128, 1599488, 32002048, 639991808, 12800032768, 255999868928, 5120000524288, 102399997902848, 2048000008388608, 40959999966445568, 819200000134217728, 16383999999463129088, 327680000002147483648, 6553599999991410065408, 131072000000034359738368

Offset: 0

Allan Bickle, Jan 02 2023

This sequence and the sequence counting the non-corner vertices (A359453) alternate as to which is larger.

			The level 1 Menger sponge graph can be formed by subdividing every edge of a cube graph.  This produces a graph with 8 corner vertices and 12 non-corner vertices, so a(1) = 8.

Allan Bickle, Degrees of Menger and Sierpinski Graphs, Congr. Num. 227 (2016) 197-208.
Allan Bickle, MegaMenger Graphs, The College Mathematics Journal, 49 1 (2018) 20-26.
Eric Weisstein's World of Mathematics, Menger Sponge
Wikipedia, Menger sponge
Index entries for linear recurrences with constant coefficients, signature (16,80).

Cf. A009964 (number of vertices), A291066 (number of edges).
Cf. A359453 (number of non-corner vertices).
Cf. A291066, A083233, and A332705 on the surface area of the n-Menger sponge graph.
Cf. A262710.

A359452[n_]:=(20^n+(-4)^n)/2;Array[A359452,25,0] (* Paolo Xausa, Nov 29 2023 *)

a(n) = (20^n + (-4)^n)/2 \\ Andrew Howroyd, Jan 02 2023

def A359452(n): return (10**n<Chai Wah Wu, Feb 13 2023

a(n) = (20^n + (-4)^n)/2.
a(n) = (A009964(n) + A262710(n))/2.
a(n) = 20^n - A359453(n).
From Stefano Spezia, Jan 02 2023: (Start)
O.g.f.: (1 - 8*x)/((1 - 20*x)*(1 + 4*x)).
E.g.f.: exp(8*x)*cosh(12*x). (End)

A359453 Number of vertices in the partite set of the n-Menger sponge graph that do not contain the corners.

Original entry on oeis.org

0, 12, 192, 4032, 79872, 1600512, 31997952, 640008192, 12799967232, 256000131072, 5119999475712, 102400002097152, 2047999991611392, 40960000033554432, 819199999865782272, 16384000000536870912, 327679999997852516352, 6553600000008589934592, 131071999999965640261632

Offset: 0

Allan Bickle, Jan 02 2023

This sequence and the sequence counting the corner vertices (A359452) alternate as to which is larger.

			The level 1 Menger sponge graph can be formed by subdividing every edge of a cube graph.  This produces a graph with 8 corner vertices and 12 non-corner vertices, so a(1) = 12.

Allan Bickle, Degrees of Menger and Sierpinski Graphs, Congr. Num. 227 (2016) 197-208.
Allan Bickle, MegaMenger Graphs, The College Mathematics Journal, 49 1 (2018) 20-26.
Eric Weisstein's World of Mathematics, Menger Sponge
Wikipedia, Menger sponge
Index entries for linear recurrences with constant coefficients, signature (16,80).

Cf. A009964 (number of vertices), A291066 (number of edges).
Cf. A359452 (number of corner vertices).
Cf. A291066, A083233, and A332705 on the surface area of the n-Menger sponge graph.

A359453[n_]:=(20^n-(-4)^n)/2;Array[A359453,25,0] (* Paolo Xausa, Nov 30 2023 *)

a(n) = (20^n - (-4)^n)/2 \\ Andrew Howroyd, Jan 02 2023

def A359453(n): return (10**n<Chai Wah Wu, Feb 13 2023

a(n) = (20^n - (-4)^n)/2.
a(n) = (A009964(n) - A262710(n))/2.
a(n) = 20^n - A359452(n).
From Stefano Spezia, Jan 02 2023: (Start)
O.g.f.: 12*x/((1 - 20*x)*(1 + 4*x)).
E.g.f.: (cosh(8*x) + sinh(8*x))*sinh(12*x). (End)

A117411 Skew triangle associated to the Euler numbers.

Original entry on oeis.org

1, 0, 1, 0, -4, 1, 0, 0, -12, 1, 0, 0, 16, -24, 1, 0, 0, 0, 80, -40, 1, 0, 0, 0, -64, 240, -60, 1, 0, 0, 0, 0, -448, 560, -84, 1, 0, 0, 0, 0, 256, -1792, 1120, -112, 1, 0, 0, 0, 0, 0, 2304, -5376, 2016, -144, 1, 0, 0, 0, 0, 0, -1024, 11520, -13440, 3360, -180, 1, 0, 0, 0, 0, 0, 0, -11264, 42240, -29568, 5280, -220, 1

Offset: 0

Paul Barry, Mar 13 2006

Inverse is A117414. Row sums of the inverse are the Euler numbers A000364.

Triangle, read by rows, given by [0,-4,4,0,0,0,0,0,0,0,...] DELTA [1,0,1,0,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938. - Philippe Deléham, Nov 01 2009

			Triangle begins
  1;
  0,  1;
  0, -4,   1;
  0,  0, -12,   1;
  0,  0,  16, -24,    1;
  0,  0,   0,  80,  -40,     1;
  0,  0,   0, -64,  240,   -60,      1;
  0,  0,   0,   0, -448,   560,    -84,      1;
  0,  0,   0,   0,  256, -1792,   1120,   -112,      1;
  0,  0,   0,   0,    0,  2304,  -5376,   2016,   -144,      1;
  0,  0,   0,   0,    0, -1024,  11520, -13440,   3360,   -180,    1;
  0,  0,   0,   0,    0,     0, -11264,  42240, -29568,   5280, -220,    1;
  0,  0,   0,   0,    0,     0,   4096, -67584, 126720, -59136, 7920, -264, 1;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Cf. A000364, A006495 (row sums), A098158, A117413, A117414.

Cf. A000217, A000332, A000579, A000581, A262710.

A117411:= func< n,k | (-4)^(n-k)*(&+[Binomial(n,k-j)*Binomial(j,n-k): j in [0..n-k]]) >;
[A117411(n,k): k in [0..n], n in [0..15]]; // G. C. Greubel, Sep 07 2022

T[n_,k_]:= T[n,k]= (-4)^(n-k)*Sum[Binomial[n, k-j]*Binomial[j, n-k], {j,0,n-k}];
Table[T[n,k], {n,0,15}, {k,0,n}]//Flatten (* G. C. Greubel, Sep 07 2022 *)

def A117411(n,k): return (-4)^(n-k)*sum(binomial(n,k-j)*binomial(j,n-k) for j in (0..n-k))
flatten([[A117411(n,k) for k in (0..n)] for n in (0..15)]) # G. C. Greubel, Sep 07 2022

Sum_{k=0..n} T(n, k) = A006495(n).
Sum_{k=0..floor(n/2)} T(n-k, k) = A117413(n).
T(n, k) = (-4)^(n-k)*Sum_{j=0..n-k} C(n,k-j)*C(j,n-k).
G.f.: (1-x*y)/(1-2x*y+x^2*y(y+4)). - Paul Barry, Mar 14 2006
T(n, k) = (-4)^(n-k)*A098158(n,k). - Philippe Deléham, Nov 01 2009
T(n, k) = 2*T(n-1,k-1) - 4*T(n-2,k-1) - T(n-2,k-2), T(0,0) = T(1,1) = 1, T(1,0) = 0, T(n,k) = 0 if k > n or if k < 0. - Philippe Deléham, Oct 31 2013
From G. C. Greubel, Sep 07 2022: (Start)
T(n, n) = 1.
T(n, n-1) = -4*A000217(n-1), n >= 1.
T(n, n-2) = (-4)^2 * A000332(n), n >= 2.
T(n, n-3) = (-4)^3 * A000579(n), n >= 3.
T(n, n-4) = (-4)^4 * A000581(n), n >= 4.
T(2*n, n) = A262710(n). (End)

A363515 Numerator of log(2) + (-1/4)^nIntegral_{x=0..1} (1 - x)^(4n+2)/(1 + x^2) dx.

Original entry on oeis.org

1, 79, 14087, 3990557, 217474889, 10326377909, 19001942777579, 3306285643032971, 3279846716611480357, 121354235196693865579, 19902098013482397470501, 1711580361934007500382731, 9009759106282339175994464009, 59689653955233943488755746919, 3820137854975012405338172218301

Offset: 0

Alexander R. Povolotsky and Stefano Spezia, Jun 07 2023

From M. F. Hasler, Jul 07 2023: (Start)
Equivalently, numerator of Sum c(n,k)/(k+1), where Sum c(n,k)*x^k = ((1 - x)^(4*n+2)/(-4)^n + 2*x)/(1 + x^2), a polynomial: The integrand (with factor (-1/4)^n) plus 2*x/(1 + x^2) is a polynomial that is easily integrated to yield the fraction a(n)/A363516(n), while Integral(-2*x/(1 + x^2)) = -log(1 + x^2) cancels the log(2).
Since the integrand/integral as a whole is less than 1/4^n in absolute value, it tends to zero and the fraction tends to log(2). (End)

			n       a(n)/A363516(n)     approximate value
-   -------------------    ------------------
0                     1       1
1                79/120       0.6583333333...
2           14087/20160       0.6987599206...
3       3990557/5765760       0.6921129218...
4   217474889/313657344       0.6933518158...
...
From _M. F. Hasler_, Jul 07 2023: (Start)
Let f[n] = (-1/4)^n*(1 - x)^(4*n+2)/(1 + x^2), the rational fraction to be integrated from 0 to 1. We have:
f[0] = 1 - 2*x/(1 + x^2), with primitive F[0] = x/2 - log(1 + x^2), whence an integral equal to 1/2 - log(2).
f[1] = -x^4/4 + (3/2)*x^3 - (7/2)*x^2 + (7/2)*x - 1/4 - 2*x/(1 + x^2), and the term-wise integration of the polynomial part gives -1/20 + 3/8 - 7/6 + 7/4 - 1/4 = 79/120, whence a(1) = 79 and A363516(1) = 120.
f[2] = (1/16)*x^8 - (5/8)*x^7 + (11/4)*x^6 - (55/8)*x^5 + (83/8)*x^4 - (71/8)*x^3 + (11/4)*x^2 + (11/8)*x + 1/16 - 2*x/(1 + x^2), so the integration gives 1/144 - 5/64 + 11/28 - 55/48 + 83/40 - 71/32 + 11/12 + 11/16 + 1/16 - log(2) = 14087/20160 - log(2), whence a(2) = 14087 and A363516(2) = 20160, etc. (End)

Mathematica Stack Exchange, How to prove using Mathematica that the sequence converges to log(2)?

Cf. A002162, A004767, A016825, A262710, A363516 (denominator).

Numerator[Simplify[Table[Log[2]+(-1)^n Integrate[(1-x)^(4n+2)/(1+x^2),{x,0,1}]/4^n,{n,0,14}]]]

A363515(n) = numerator(subst(intformal(((1-x)^(4*n+2)/(-4)^n+2*x)/(1+x^2)),x,1)) \\ The argument of intformal is a polynomial that is trivially integrated over [0, 1]. - M. F. Hasler, Jul 07 2023

Numerator of log(2) + HypergeometricPFQ([1/2, 1, 1], [2*(1 + n), 5/2 + 2*n], -1)/((3 + 4*n)*(-4)^n).

Limit_{n->oo} a(n)/A363516(n) = log(2).

A363516 Denominator of log(2) + (-1/4)^nIntegral_{x=0..1} (1 - x)^(4n+2)/(1 + x^2) dx.

Original entry on oeis.org

1, 120, 20160, 5765760, 313657344, 14898723840, 27413651865600, 4769975424614400, 4731815621217484800, 175077177985046937600, 28712657189547697766400, 2469288518301102007910400, 12998334760337000969640345600, 86113967787232631423867289600, 5511293938382888411127506534400

Offset: 0

Alexander R. Povolotsky and Stefano Spezia, Jun 07 2023

Equivalently, denominator of Sum c(n,k)/(k+1), where ((1 - x)^(4*n+2)/(-4)^n + 2*x)/(1 + x^2) = Sum c(n,k)*x^k, a polynomial. In other words, the integrand (with factor (-1/4)^n) plus 2*x/(1 + x^2) is a polynomial that can easily be term-wise integrated to yield the fraction A363515(n)/a(n), while antiderivative(-2*x/(1 + x^2)) = -log(1 + x^2) cancels the log(2). - M. F. Hasler, Jul 07 2023

			n       A363515(n)/a(n)    approximate value
-   -------------------    -----------------
0                     1      1
1                79/120      0.6583333333...
2           14087/20160      0.6987599206...
3       3990557/5765760      0.6921129218...
4   217474889/313657344      0.6933518158...
...
From _M. F. Hasler_, Jul 07 2023: (Start)
Let f[n] = (-1/4)^n*(1 - x)^(4*n+2)/(1 + x^2), the rational fraction to be integrated from 0 to 1. We have:
f[0] = 1 - 2*x/(1 + x^2), with primitive F[0] = x/2 - log(1 + x^2), whence an integral equal to 1/2 - log(2), and a(0) = 2 (denominator).
f[1] = -x^4/4 + (3/2)*x^3 - (7/2)*x^2 + (7/2)*x - 1/4 - 2*x/(1 + x^2), and the term-wise integration of the polynomial part gives -1/20 + 3/8 - 7/6 + 7/4 - 1/4 = 79/120, whence A363515(1) = 79 and a(1) = 120.
f[2] = (1/16)*x^8 - (5/8)*x^7 + (11/4)*x^6 - (55/8)*x^5 + (83/8)*x^4 - (71/8)*x^3 + (11/4)*x^2 + (11/8)*x + 1/16 - 2*x/(1 + x^2), so the integration gives 1/144 - 5/64 + 11/28 - 55/48 + 83/40 - 71/32 + 11/12 + 11/16 + 1/16 - log(2) = 14087/20160 - log(2), whence A363515(2) = 14087 and a(2) = 20160, etc. (End)

Mathematica Stack Exchange, How to prove using Mathematica that the sequence converges to log(2)?

Cf. A002162, A004767, A016825, A262710, A363515 (numerator).

Denominator[Simplify[Table[Log[2]+(-1)^n Integrate[(1-x)^(4n+2)/(1+x^2),{x,0,1}]/4^n,{n,0,14}]]]

Denominator of log(2) + HypergeometricPFQ([1/2, 1, 1], [2*(1 + n), 5/2 + 2*n], -1)/((3 + 4*n)*(-4)^n).

Limit_{n->oo} A363515(n)/a(n) = log(2).

A350384 a(n) = (-1728)^n.

Original entry on oeis.org

1, -1728, 2985984, -5159780352, 8916100448256, -15407021574586368, 26623333280885243904, -46005119909369701466112, 79496847203390844133441536, -137370551967459378662586974208, 237376313799769806328950291431424, -410186270246002225336426103593500672

Offset: 0

Stefano Spezia, Dec 28 2021

Caroline Nunn, A Proof of a Generalization of Niven's Theorem Using Algebraic Number Theory, Rose-Hulman Undergraduate Mathematics Journal: Vol. 22, Iss. 2, Article 3 (2021).
Index entries for linear recurrences with constant coefficients, signature (-1728).

Cf. A000244, A000578, A001021, A008585, A008588, A262710.

LinearRecurrence[{-1728},{1},12]
NestList[-1728#&,1,20] (* Harvey P. Dale, Dec 25 2023 *)

From Caroline Nunn, p. 9: (Start)
a(n) = (3 + sqrt(-3))^(6*n).
a(n) = Sum_{k=0..3*n} (-1)^k*binomial(6*n, 2*k)*3^(6*n-k). (End)
O.g.f.: 1/(1 + 1728*x).
E.g.f.: exp(-1728*x).
a(n) = -1728*a(n-1) for n > 0.
a(n) = (-12)^(3*n).
a(n) = (A000244(n)*A262710(n))^3.

cp's OEIS Frontend

A359452 Number of vertices in the partite set of the n-Menger sponge graph that contains the corners.

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A359453 Number of vertices in the partite set of the n-Menger sponge graph that do not contain the corners.

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A117411 Skew triangle associated to the Euler numbers.

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A363515 Numerator of log(2) + (-1/4)^n*Integral_{x=0..1} (1 - x)^(4*n+2)/(1 + x^2) dx.

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A363516 Denominator of log(2) + (-1/4)^n*Integral_{x=0..1} (1 - x)^(4*n+2)/(1 + x^2) dx.

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A350384 a(n) = (-1728)^n.

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A363515 Numerator of log(2) + (-1/4)^nIntegral_{x=0..1} (1 - x)^(4n+2)/(1 + x^2) dx.

A363516 Denominator of log(2) + (-1/4)^nIntegral_{x=0..1} (1 - x)^(4n+2)/(1 + x^2) dx.