A114239 -id:A114239 - cp's OEIS frontend

A107891 a(n) = (n+1)(n+2)^2(n+3)^2(n+4)(3n^2 + 15n + 20)/2880.

1, 19, 155, 805, 3136, 9996, 27468, 67320, 150645, 313027, 611611, 1134497, 2012920, 3436720, 5673648, 9093096, 14194881, 21643755, 32310355, 47319349, 68105576, 96479020, 134699500, 185562000, 252493605, 339663051, 452103939

Offset: 0

Emeric Deutsch, Jun 12 2005

Kekulé numbers for certain benzenoids.

Partial sums of A114239. First differences of A047819. - Peter Bala, Sep 21 2007

S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (see pp. 167, 187 and p. 105 eq. (iii) for k=2 and m=5).

Shawn A. Broyles, Table of n, a(n) for n = 0..1000
Paolo Aluffi, Degrees of projections of rank loci, arXiv:1408.1702 [math.AG], 2014. ["After compiling the results of many explicit computations, we noticed that many of the numbers d_{n,r,S} appear in the existing literature in contexts far removed from the enumerative geometry of rank conditions; we owe this surprising (to us) observation to perusal of [Slo14]."]
Index entries for linear recurrences with constant coefficients, signature (9,-36,84,-126,126,-84,36,-9,1).

Cf. A005585, A006542, A047819, A114239, A114242.

a:=n->(1/2880)*(n+1)*(n+2)^2*(n+3)^2*(n+4)*(3*n^2+15*n+20): seq(a(n),n=0..32);

Table[((1+n) (2+n)^2 (3+n)^2 (4+n) (20+3 n (5+n)))/2880,{n,0,40}] (* or *) LinearRecurrence[{9,-36,84,-126,126,-84,36,-9,1},{1,19,155,805,3136,9996,27468,67320,150645},40] (* Harvey P. Dale, Dec 10 2021 *)

a(n-2) = (1/8) * Sum_{1 <= x_1, x_2 <= n} (x_1*x_2)^2*(det V(x_1,x_2))^2 = 1/8*sum {1 <= i,j <= n} (i*j*(i-j))^2, where V(x_1,x_2) is the Vandermonde matrix of order 2. - Peter Bala, Sep 21 2007
G.f.: (1+10*x+20*x^2+10*x^3+x^4)/(1-x)^9. - Colin Barker, Feb 08 2012
a(n) = (A000330(n+2)*A000538(n+2) - (A000537(n+2))^2)/4. - J. M. Bergot, Sep 17 2013
Sum_{n>=0} 1/a(n) = 17095/4 - 240*Pi^2 - 162*sqrt(15)*Pi*tanh(sqrt(5/3)*Pi/2). - Amiram Eldar, May 29 2022

A195166 Numbers expressible as 2^a - 2^b, with 0 <= b < a, such that n^a - n^b is divisible by 2^a - 2^b for all n.

Original entry on oeis.org

1, 2, 6, 12, 30, 24, 60, 120, 252, 240, 504, 16380, 32760, 65520

Offset: 1

Michel Marcus, Dec 21 2012

= 2^1  - 2^0. (n^1  - n^0)/1     : A000027
= 2^2  - 2^1. (n^2  - n^1)/2     : A000217
= 2^3  - 2^1. (n^3  - n^1)/6     : A000292
= 2^4  - 2^2. (n^4  - n^2)/12    : A002415
= 2^5  - 2^1. (n^5  - n^1)/30    : A033455
= 2^5  - 2^3. (n^5  - n^3)/24    : A006414
= 2^6  - 2^2. (n^6  - n^2)/60    : A213547
= 2^7  - 2^3. (n^7  - n^3)/120   : A114239
= 2^8  - 2^2. (n^8  - n^2)/252   :
= 2^8  - 2^4. (n^8  - n^4)/240   : A078876
= 2^9  - 2^3. (n^9  - n^3)/504   :
= 2^14 - 2^2. (n^14 - n^2)/16380 :
= 2^15 - 2^3. (n^15 - n^3)/32760 :
= 2^16 - 2^4. (n^16 - n^4)/65520 :
Comment from Tomohiro Yamada, Oct 05 2022: (Start)
"The Mod Set Stanford University" and Carl Pomerance independently noted that the completeness of this sequence follows from a result of Schinzel on primitive prime factors of sequences a^n - b^n in the remark to a problem of Harry Ruderman asking whether if 2^a - 2^b divides 3^a - 3^b, then 2^a - 2^b belongs to this sequence.
The Mod Set verified that if a > b, 2^a - 2^b divides 3^a - 3^b, but 2^a - 2^b does not belong to this sequence, then a - b > 1900. Ram Murty and Kumar Murty proved that there are only finitely many natural numbers a, b such that 2^a - 2^b divides 3^a - 3^b.
Qi Sun and Ming Zhi Zhang also showed that if a > b and n^a - n^b is divisible by 2^a - 2^b for all n, then 2^a - 2^b belongs to this sequence. (End)

			a(3) = 6 belongs to this sequence since (n^3 - n)/6 = C(n+1, 3) = A000292(n-1).

M. Ram Murty and V. Kumar Murty, On a Problem of Ruderman, Amer. Math. Monthly 118 (2011), 644-650, available from the first author's website.
Harry Ruderman, Problem E2468, Amer. Math. Monthly 81 (1974), p. 405.
A. Schinzel, On primitive prime factors of a^n - b^n, Proc. Cambridge Phil. Soc. 58 (1962), 556-562.
Qi Sun and Ming Zhi Zhang, Pairs where 2^a-2^b divides n^a-n^b for all n, Proc. Amer. Math. Soc. 93 (1985), 218-220.
The Mod Set Stanford University and Carl Pomerance, When 2^m - 2^n divides 3^m - 3^n, remarks to Problem E2468*, Amer. Math. Monthly 84 (1977), 59-60.
W. Y. Velez, When 2^m - 2^n divides 3^m - 3^n, remarks to Problem E2468, Amer. Math. Monthly 83 (1976), 288-289.

cp's OEIS Frontend

A107891 a(n) = (n+1)(n+2)^2(n+3)^2(n+4)(3n^2 + 15n + 20)/2880.

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A195166 Numbers expressible as 2^a - 2^b, with 0 <= b < a, such that n^a - n^b is divisible by 2^a - 2^b for all n.

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cp's OEIS Frontend

A107891 a(n) = (n+1)*(n+2)^2*(n+3)^2*(n+4)*(3n^2 + 15n + 20)/2880.

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Maple

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A195166 Numbers expressible as 2^a - 2^b, with 0 <= b < a, such that n^a - n^b is divisible by 2^a - 2^b for all n.

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A107891 a(n) = (n+1)(n+2)^2(n+3)^2(n+4)(3n^2 + 15n + 20)/2880.