Melvin Peralta - cp's OEIS frontend

A295319 a(n) is the sum of all n-digit palindromes.

45, 495, 49500, 495000, 49500000, 495000000, 49500000000, 495000000000, 49500000000000, 495000000000000, 49500000000000000, 495000000000000000, 49500000000000000000, 495000000000000000000, 49500000000000000000000, 495000000000000000000000

Offset: 1

Melvin Peralta, Nov 19 2017

For n > 1, the sum of the digits of a(n) is always 18 (see AoPS link).

Conjecture: All terms after a(1) are of the form 495*10^x where x is any nonnegative integer k such that k != 1 (mod 3). - Harvey P. Dale, Mar 05 2023

			The sum of all nine two-digit palindromes is 11 + 22 + 33 + 44 + 55 + 66 + 77 + 88 + 99 = 495, and so a(2) = 495.
The sum of all three-digit palindromes is (101 + 999) + (111 + 989) + (121 + 979) + ... (545 + 565) + 555 = 49500, and so a(3) = 49500.

AoPS, problem 15 of the 2014 American Mathematics Competition 12
Index entries for linear recurrences with constant coefficients, signature (0,1000).

Cf. A002113, A050683, A261675.

palSum[n_] := 99/2*10^(n - 1) * 10^Floor[(n - 1)/2]; palSum[1] = 45; Array[ palSum, 16] (* Robert G. Wilson v, Nov 21 2017 *)

a(n) = if (n==1, 45, 9*10^floor((n-1)/2)*11*10^(n-1)/2); \\ Michel Marcus, Dec 26 2017

a(n) = A050683(n)*(5*10^(n-1) + (9/2)*10^(n-2) + ... + (9/2)*10 + 5) (calculates the sum by multiplying the expected value of a randomly selected n-digit palindrome with the number of n-digit palindromes).
For n > 1, a(n) = (A050683(n)/2)*11*10^(n-1).
For n > 3, a(n) = 1000 * a(n - 2). - David A. Corneth, Dec 26 2017
G.f.: x*(45 + 495*x + 4500*x^2)/(1 - 1000*x^2). - Chai Wah Wu, Jan 22 2018
E.g.f.: 9*(11*cosh(10*sqrt(10)*x) + 11*sqrt(10)*sinh(10*sqrt(10)*x) - 11 - 100*x)/200. - Stefano Spezia, Sep 19 2024

A286899 Array read by antidiagonals: A(n, L) is the number of closed walks of length 2L along the edges of an n-cube based at a vertex, for n >= 1 and L >= 1.

Original entry on oeis.org

1, 1, 2, 1, 8, 3, 1, 32, 21, 4, 1, 128, 183, 40, 5, 1, 512, 1641, 544, 65, 6, 1, 2048, 14763, 8320, 1205, 96, 7, 1, 8192, 132861, 131584, 26465, 2256, 133, 8, 1, 32768, 1195743, 2099200, 628805, 64896, 3787, 176, 9, 1, 131072, 10761681, 33562624, 15424865

Offset: 1

Melvin Peralta, May 15 2017

			A(2, 2) = 8 because at each vertex of a 2-cube (i.e., a square), there are 8 closed walks of length 2(2) = 4.
A(1, k) = 1 because at the vertex of a 1-cube, there is 1 closed walk of any length 2*k.
Array A(n, L) begins:
   1         1         1         1         1         1 ...
   2         8        32       128       512      2048 ...
   3        21       183      1641     14763    132861 ...
   4        40       544      8320    131584   2099200 ...
   5        65      1205     26465    628805  15424865 ...
   6        96      2256     64896   2086656  71172096 ...
   7       133      3787    134953   5501167 243147373 ...

R. P. Stanley, Algebraic Combinatorics: Walks, Trees, Tableaux, and More, Springer, 2013.

Melvin Peralta, Table of n, a(n) for n = 1..120

Cf. A054879, A092812, A121822.

A286899 := proc(n,L)
    add(binomial(n,i)*(n-2*i)^L, i=0..n) ;
    %/2^n ;
end proc:
for n from 1 to 7 do
    for L from 2 to 12 by 2 do
        printf("%9d ",A286899(n,L)) ;
    end do:
    printf("\n") ;
end do: # R. J. Mathar, May 22 2017

f[n_, l_] := 1/2^n*Sum[Binomial[n, i]*(n - 2 i)^l, {i, 0, n}];
Table[f[n - l + 1, 2 l], {n, 1, 15}, {l, n, 1, -1}] // Flatten

A(n, L) = (1/2^n)*Sum_{i=0..n} binomial(n, i)*(n - 2*i)^(2*L). (Corrected by Peter Luschny, Jul 07 2019.)

A285079 Oblong numbers the product of whose digits are positive oblong numbers.

Original entry on oeis.org

2, 6, 12, 56, 132, 156, 756, 2756, 4556, 6162, 6972, 7656, 13572, 21756, 31152, 33672, 45156, 61752, 84972, 153272, 166872, 279312, 467172, 626472, 661782, 1273512, 1412532, 1541322, 1568756, 1596432, 1786232, 1867322, 2678132, 2817362, 3416952, 3521252

Offset: 1

Melvin Peralta, Apr 09 2017

Oblong numbers are numbers of the form k*(k+1) (A002378).

Robert Israel and Giovanni Resta, Table of n, a(n) for n = 1..391 (terms < 10^25, first 221 terms from Robert Israel)

Cf. A002378, A053059, A069077.

filter:= proc(x) local t; t:= convert(convert(x,base,10),`*`);
t > 0 and issqr(1+4*t); end proc:
select(filter, [seq(x*(1+x),x=1..10^4)]); # Robert Israel, Apr 14 2017

f[x_] := Sqrt[1 + 4 (Times @@ IntegerDigits[x])]; Select[Table[n (n + 1), {n, 1, 10000}], f[#] > 1 && Mod[f[#], 2] == 1 &]

A272934 Depth of Pascal's triangle such that the number of elements in the triangle is a factor of the sum of the elements.

Original entry on oeis.org

1, 2, 6, 18, 42, 126, 162, 378, 486, 882, 1458, 2646, 3078, 3942, 5418, 9198, 11826, 14406, 16758, 18522, 24966, 26406, 37338, 39366, 42462, 71442, 77658, 95922, 99078, 113778, 117306, 143262, 174762, 175446, 184842, 265482, 304038, 308826, 318402, 351918

Offset: 1

Melvin Peralta, May 11 2016

nonn

a(n) are the values m such that the expression (2^(m+1) - 2)/(m^2 + m) is an integer.
a(n) are the values m such that A000225(m)/A000217(m) is an integer.
It appears that a(n) == 2 (mod 4) for n > 1. - Robert Israel, Jul 04 2017

			a(2) = 6 because if Pascal's triangle is written out to 6 rows, there will be 21 elements whose sum is 63, and 21 is a factor of 63.
6 is a term because A000225(6)/A000217(6) = 63/21 = 3, an integer.

Robert Israel, Table of n, a(n) for n = 1..300 (first 66 terms from Melvin Peralta)

Cf. A000217, A000225, A007318.

select(t -> 2 &^ t  - 1 mod t*(t+1)/2 = 0, [$1..10^6]); # Robert Israel, Jul 04 2017

Join[{1}, Select[Range[10^6], PowerMod[2, #+1, #^2+#] == 2 &]]

Mild editing. Wolfdieter Lang, May 31 2016

A268283 Number of distinct directed Hamiltonian cycles of the Platonic graphs (in the order of tetrahedral, cubical, octahedral, dodecahedral, and icosahedral graph).

Original entry on oeis.org

6, 12, 32, 60, 2560

Offset: 1

Melvin Peralta, Jan 29 2016

a(n)/2 is the number of distinct undirected Hamiltonian cycles of the Platonic graph corresponding to a(n).

Eric Weisstein's World of Mathematics, Tetrahedral Graph
Eric Weisstein's World of Mathematics, Cubical Graph
Eric Weisstein's World of Mathematics, Octahedral Graph
Eric Weisstein's World of Mathematics, Dodecahedral Graph
Eric Weisstein's World of Mathematics, Icosahedral Graph

Cf. A052762 (tetrahedral graph), A140986 (cubical graph), A115400 (octahedral graph), A218513 (dodecahedral graph), A218514 (icosahedral graph).

A268135 Numbers n such that the digit sum of n^2 is a divisor of the digit sum of n.

Original entry on oeis.org

1, 9, 10, 18, 19, 45, 46, 55, 90, 99, 100, 145, 149, 180, 189, 190, 198, 199, 289, 351, 361, 369, 379, 388, 450, 451, 459, 460, 468, 495, 496, 549, 550, 558, 559, 568, 585, 595, 639, 729, 739, 775, 838, 855, 900, 954, 955, 990, 999, 1000, 1049, 1098, 1099, 1179, 1188, 1189, 1198

Offset: 1

Melvin Peralta, Jan 26 2016

Because A058369 (with offset 1) is a subsequence, this sequence is infinite.
Conjecture: The relative complement of A058369 with respect to this sequence is infinite. That is, there are infinitely many n such that the digit sum of n^2 is a proper divisor of the digit sum of n.
If the digit sum of n^2 is a proper divisor of the digit sum of n, then this property holds for 10*n as well, i.e. the digit sum of n = 149*10^k has as a proper divisor the digit sum of n^2 for all k > 0. Are there infinitely many n that are not a multiple of 10 such that the digit sum of n^2 is a proper divisor of the digit sum of n? The first few such numbers are: 149, 549, 1049, 14499, 19499, 55679, 59499, 64499, 73499, 118499, 144999, 145949, 179249, 244949, 244998, 334679, 347855, 473499, 548735, 549549, 549639, 556965, 837855, ... - Chai Wah Wu, Mar 16 2016

			Digit sum of 149^2 = 7. Digit sum of 149 = 14. Since 7 is a divisor of 14, 149 is in the sequence.

Indranil Ghosh, Table of n, a(n) for n = 1..10001

Cf. A007953, A004159, A058369.

Select[Range[200], Mod[Total[IntegerDigits[#]], Total[IntegerDigits[#^2]]] == 0 &]

isok(n) = (sumdigits(n) % sumdigits(n^2)) == 0; \\ Michel Marcus, Jan 27 2016

More terms from Michel Marcus, Jan 27 2016

A265853 a(0)=1; for n >= 1, a(n) is the number of subsets of [a(0), a(1), ..., a(n-1)] whose sum is equal to a(n-1).

Original entry on oeis.org

1, 1, 2, 2, 3, 5, 6, 8, 11, 17, 25, 33, 41, 52, 80, 139, 204, 245, 289, 410, 692, 1159, 1477, 2010, 2769, 4247, 6128, 7709, 9817, 14071, 21982, 34892, 52079, 63998, 81167, 122709, 183662, 267520, 382690, 521361, 725601, 1050579, 1541163, 2084690, 2829408

Offset: 0

Melvin Peralta, Dec 21 2015

nonn

			a(4) = 3 because there are subsets of [1, 1, 2, 2] that sum to a(3) = 2: {1, 1}, {2}, {2}.

Alois P. Heinz, Table of n, a(n) for n = 0..85 (first 71 terms from Bert Dobbelaere)

Cf. A057601.

s:= proc(n) option remember; `if`(n<0, 0, s(n-1)+a(n)) end:
b:= proc(n, i) option remember; `if`(n=0, 1, `if`(s(i) `if`(p>=0, b(p, i-1), 0))(n-a(i))))
    end:
a:= n-> `if`(n=0, 1, b(a(n-1), n-1)):
seq(a(n), n=0..44);  # Alois P. Heinz, Jan 24 2024

lst={1};n=1;While[n<30, lst = Join[lst, {Length@Select[Total /@ Subsets[lst],#==Last[lst]&]}];n++]

a(0) = 1; a(n) = [x^a(n-1)] Product_{k=0..n-1} (1 + x^a(k)). - Ilya Gutkovskiy, Jan 24 2024

More terms from Bert Dobbelaere, Oct 28 2018

A268962 a(1) = 1 and a(2) = 2; thereafter a(n+1) = floor[sqrt(a(n))] if not already in the sequence; otherwise a(n+1) = a(n) + a(n-1).

Original entry on oeis.org

1, 2, 3, 5, 8, 13, 21, 4, 25, 29, 54, 7, 61, 68, 129, 11, 140, 151, 12, 163, 175, 338, 18, 356, 374, 19, 393, 412, 20, 432, 452, 884, 1336, 36, 6, 42, 48, 90, 9, 99, 108, 10, 118, 128, 246, 15, 261, 16, 277, 293, 17, 310, 327, 637, 964, 31, 995, 1026, 32

Offset: 1

Melvin Peralta, Feb 16 2016

nonn

Hans Havermann found that the sequence is not a permutation of the integers since a(708)=a(1276)=1666.

Melvin Peralta, Table of n, a(n) for n = 1..20000

Cf. A014682, A114183, A000045

seq = {1, 2}; max; n = 3; While[n < max, If[MemberQ[seq, Floor[Sqrt[Last[seq]]]] == False, AppendTo[seq, Floor[Sqrt[Last[seq]]]], AppendTo[seq, Last[seq] + Part[seq, n - 2]]]; n++]

A268962_vec=A268962_used=[1,2]; A268962(n)={ #A268962_vec>=n || A268962_vec=concat(A268962_vec,vector(n-#A268962_vec)); A268962_vec[n] || for(n=#A268962_used+A268962_used[1],n, A268962_vec[n]&&next; (A268962_vec[n-1]<(A268962_used[1]+1)^2 || setsearch(A268962_used,A268962_vec[n]=sqrtint(A268962_vec[n-1]))) && A268962_vec[n]=A268962_vec[n-1]+A268962_vec[n-2]; A268962_used=setunion(A268962_used,A268962_vec[n..n]); while(#A268962_used>1 && A268962_used[2]==A268962_used[1]+1,A268962_used=A268962_used[^1])); A268962_vec[n]} \\ M. F. Hasler, Feb 16 2016

cp's OEIS Frontend

User: Melvin Peralta

A295319 a(n) is the sum of all n-digit palindromes.

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A286899 Array read by antidiagonals: A(n, L) is the number of closed walks of length 2L along the edges of an n-cube based at a vertex, for n >= 1 and L >= 1.

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A285079 Oblong numbers the product of whose digits are positive oblong numbers.

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A272934 Depth of Pascal's triangle such that the number of elements in the triangle is a factor of the sum of the elements.

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A268283 Number of distinct directed Hamiltonian cycles of the Platonic graphs (in the order of tetrahedral, cubical, octahedral, dodecahedral, and icosahedral graph).

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A268135 Numbers n such that the digit sum of n^2 is a divisor of the digit sum of n.

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Mathematica

PARI

Extensions

A265853 a(0)=1; for n >= 1, a(n) is the number of subsets of [a(0), a(1), ..., a(n-1)] whose sum is equal to a(n-1).

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A268962 a(1) = 1 and a(2) = 2; thereafter a(n+1) = floor[sqrt(a(n))] if not already in the sequence; otherwise a(n+1) = a(n) + a(n-1).