A191681 -id:A191681 - cp's OEIS frontend

A256786 Numbers which are divisible by prime(d) for all digits d in their decimal representation.

12, 14, 42, 55, 154, 222, 228, 714, 1122, 1196, 1212, 1414, 2112, 2142, 2262, 3355, 4144, 4242, 5335, 5544, 5555, 6162, 9499, 11112, 11144, 11214, 11424, 11466, 11622, 11818, 11914, 12222, 12882, 14112, 15554, 16666, 21216, 21222, 21252, 21888, 22122, 22212

Offset: 1

Eric Angelini and Reinhard Zumkeller, Apr 10 2015

All terms are zerofree, cf. A052382;
there is no term containing digits 1 and 3 simultaneously;
a(n) contains at least one digit 1 iff a(n) is even, cf. A011531, A005843;
a(n) contains at least one digit 2 iff a(n) mod 3 = 0, cf. A011532, A008585;
a(n) contains at least one digit 3 iff a(n) mod 10 = 5, cf. A011533, A017329;
A020639(a(n)) <= 23.
The equivalent in base 2 is the empty sequence, in base 3 it is A191681\{0}; see A256874-A256879 for the base 4 - base 9 variant, and A256870 for a variant where digits 0 are allowed but divisibility by prime(d+1) is required instead. - M. F. Hasler, Apr 11 2015

			Smallest terms containing the nonzero decimal digits:
.  d | prime(d) |  n | a(n)
. ---+----------+--------------------------
.  1 |       2  |  1 |   12 = 2^2 * 3
.  2 |       3  |  1 |   12 = 2^2 * 3
.  3 |       5  | 16 | 3355 = 5 * 11 * 61
.  4 |       7  |  2 |   14 = 2 * 7
.  5 |      11  |  4 |   55 = 5 * 11
.  6 |      13  | 10 | 1196 = 2^2 * 13 * 23
.  7 |      17  |  8 |  714 = 2 * 3 * 7 * 17
.  8 |      19  |  7 |  228 = 2^2 * 3 * 19
.  9 |      23  | 10 | 1196 = 2^2 * 13 * 23 .

Lars Blomberg and Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Éric Angelini, Divisibility by primes, SeqFan list, Apr 10 2015.
Index entries for 10-automatic sequences.

Cf. A000040, A005843, A008585, A011531, A011532, A011533, A017329, A020639, A052382, A256874-A256879, A256882-A256884, A256865-A256870.

a256786 n = a256786_list !! (n-1)
a256786_list = filter f a052382_list where
   f x = g x where
     g z = z == 0 || x `mod` a000040 d == 0 && g z'
           where (z', d) = divMod z 10

Select[Range@22222,FreeQ[IntegerDigits[#],0]&&Total[Mod[#,Prime[IntegerDigits[#]]]]==0&] (* Ivan N. Ianakiev, Apr 11 2015 *)

is_A256786(n)=!for(i=1,#d=Set(digits(n)),(!d[i]||n%prime(d[i]))&&return) \\ M. F. Hasler, Apr 11 2015

primes = [1, 2, 3, 5, 7, 11, 13, 17, 19, 23]
def ok(n):
    s = str(n)
    return "0" not in s and all(n%primes[int(d)] == 0 for d in s)
print([k for k in range(22213) if ok(k)]) # Michael S. Branicky, Dec 14 2021

A125857 Numbers whose base-9 representation is 22222222.......2.

Original entry on oeis.org

0, 2, 20, 182, 1640, 14762, 132860, 1195742, 10761680, 96855122, 871696100, 7845264902, 70607384120, 635466457082, 5719198113740, 51472783023662, 463255047212960, 4169295424916642, 37523658824249780, 337712929418248022

Offset: 1

Zerinvary Lajos, Feb 03 2007

If f(1) := 1/x and f(n+1) = (f(n) + 2/f(n))/3, then f(n) = 3^(1-n) * (1/x + a(n)*x + O(x^3)). - Michael Somos, Jul 28 2020

			G.f. = 2*x^2 + 20*x^3 + 182*x^4 + 1640*x^5 + 14762*x^6 + 132860*x^7 + ... - _Michael Somos_, Jul 28 2020

G. Benkart, D. Moon, A Schur-Weyl Duality Approach to Walking on Cubes, arXiv preprint arXiv:1409.8154 [math.RT], 2014 and Ann. Combin. 20 (3) (2016) 397-417
E. Estrada and J. A. de la Pena, From Integer Sequences to Block Designs via Counting Walks in Graphs, arXiv preprint arXiv:1302.1176 [math.CO], 2013. - From _N. J. A. Sloane_, Feb 28 2013
E. Estrada and J. A. de la Pena, Integer sequences from walks in graphs, Notes on Number Theory and Discrete Mathematics, Vol. 19, 2013, No. 3, 78-84.
R. J. Mathar, Counting Walks on Finite Graphs, Nov 2020, Section 5.
Vladimir Pletser, Congruence conditions on the number of terms in sums of consecutive squared integers equal to squared integers, arXiv:1409.7969 [math.NT], 2014.
Index entries for linear recurrences with constant coefficients, signature (10,-9).

Cf. A002452.

Maple
```
seq((9^n-1)*2/8, n=0..19);
```

FromDigits[#, 9]&/@Table[PadRight[{2}, n, 2], {n, 0, 20}] (* Harvey P. Dale, Feb 02 2011 *)
Table[(9^(n - 1) - 1)*2/8, {n, 20}] (* Wesley Ivan Hurt, Mar 29 2014 *)

Vec(2*x^2/((x-1)*(9*x-1)) + O(x^100)) \\ Colin Barker, Sep 30 2014

{a(n) = (9^(n-1) - 1)/4}; /* Michael Somos, Jul 02 2017 */

a(n) = (9^(n-1) - 1)*2/8.
a(n) = 9*a(n-1) + 2 (with a(1)=0). - Vincenzo Librandi, Sep 30 2010
a(n) = 2 * A002452(n). - Vladimir Pletser, Mar 29 2014
From Colin Barker, Sep 30 2014: (Start)
a(n) = 10*a(n-1) - 9*a(n-2).
G.f.: 2*x^2 / ((x-1)*(9*x-1)). (End)
a(n) = -a(2-n) * 9^(n-1) for all n in Z. - Michael Somos, Jul 02 2017
a(n) = A191681(n-1)/2. - Klaus Purath, Jul 03 2020

A350993 Triangular numbers that are palindromes in base 9.

Original entry on oeis.org

0, 1, 3, 6, 10, 91, 136, 300, 528, 820, 4560, 7381, 11476, 20910, 42486, 66430, 552826, 581581, 597871, 1664400, 2001000, 3420420, 3444000, 5070520, 5380840, 48427561, 75995956, 132494781, 134553810, 137158203, 159213090, 290585778, 434520460, 435848050, 669615310

Offset: 1

Amiram Eldar, Jan 28 2022

This sequence is infinite since A000217((9^k-1)/2) is a term for all k >= 0 (Wishard, 1931).

Also, A000217((3 + 5*9^k)/2) is a term for all k>=0 (Trigg, 1984).

			10 is a term since 10 = A000217(4) is a triangular number and also a palindromic number in base 9: 10 = 11_9.
91 is a term since 91 = A000217(13) is a triangular number and also a palindromic number in base 9: 91 = 111_9.

Charles W. Trigg, Mathematical Quickies, McGraw Hill Book Co., 1967, Q112, p. 127.

Amiram Eldar, Table of n, a(n) for n = 1..123
Charles W. Trigg, Infinite sequences of palindromic triangular numbers, The Fibonacci Quarterly, Vol. 12, No. 2 (1974), pp. 209-212.
Charles W. Trigg, Problem 281, The College Mathematics Journal, Vol. 15, No. 4 (1984), p. 346; Palindromic Triangular Numbers in Base Nine, Solution to Problem 281, by Michael Vowe, ibid., Vol. 17, No. 2 (1986), pp. 188-189.
Maciej Ulas, On certain diophantine equations related to triangular and tetrahedral numbers, arXiv:0811.2477 [math.NT], 2008.
G. W. Wishard, Problem 3480, The American Mathematical Monthly, Vol. 38, No. 3 (1931), p. 170; Solution to Problem 3480, by Helen A. Merrill, ibid., Vol. 39, No. 3 (1932), p. 179.

Intersection of A000217 and A029955.
The nonary version of A003098.
Cf. A191681, A350987, A350990, A350991, A350992.

t[n_] := n*(n + 1)/2; Select[t /@ Range[0, 3*10^5], PalindromeQ[IntegerDigits[#, 9]] &]

A255043 a(n) = (5*9^n - 1)/2.

Original entry on oeis.org

2, 22, 202, 1822, 16402, 147622, 1328602, 11957422, 107616802, 968551222, 8716961002, 78452649022, 706073841202, 6354664570822, 57191981137402, 514727830236622, 4632550472129602, 41692954249166422, 375236588242497802, 3377129294182480222

Offset: 0

L. Edson Jeffery, Feb 13 2015

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (10,-9).

Cf. A096053, A138894, A191681, A198964, A198969.

Row 2 of A255044.

[(5*9^n -1)/2: n in [0..20]]; // G. C. Greubel, Feb 07 2021

Table[(5*9^n - 1)/2, {n, 0, 19}]
LinearRecurrence[{10,-9},{2,22},20] (* Harvey P. Dale, Jun 15 2018 *)

[(5*9^n -1)/2 for n in (0..20)] # G. C. Greubel, Feb 07 2021

G.f.: 2*(1+x)/((1-x)*(1-9*x)).
Recurrence: a(n) = 10*a(n-1) - 9*a(n-2), n>=2, a(0) = 2, a(1) = 22.
a(n) = 2*A138894(n).
E.g.f.: (5*exp(9*x) - exp(x))/2. - G. C. Greubel, Feb 07 2021

A255044 Array A read by upward antidiagonals: A(n,k) = ((2n+1)9^k-1)/2, n,k >= 0.

Original entry on oeis.org

0, 1, 4, 2, 13, 40, 3, 22, 121, 364, 4, 31, 202, 1093, 3280, 5, 40, 283, 1822, 9841, 29524, 6, 49, 364, 2551, 16402, 88573, 265720, 7, 58, 445, 3280, 22963, 147622, 797161, 2391484, 8, 67, 526, 4009, 29524, 206671, 1328602, 7174453, 21523360

Offset: 0

L. Edson Jeffery, Feb 13 2015

			Array begins:
.   0   4   40   364   3280   29524   265720   2391484   21523360
.   1  13  121  1093   9841   88573   797161   7174453   64570081
.   2  22  202  1822  16402  147622  1328602  11957422  107616802
.   3  31  283  2551  22963  206671  1860043  16740391  150663523
.   4  40  364  3280  29524  265720  2391484  21523360  193710244
.   5  49  445  4009  36085  324769  2922925  26306329  236756965
.   6  58  526  4738  42646  383818  3454366  31089298  279803686
.   7  67  607  5467  49207  442867  3985807  35872267  322850407
.   8  76  688  6196  55768  501916  4517248  40655236  365897128

Cf. A191681, A096053, A255043, A198964, A198969 (rows 0-3 and 5).

Cf. A138894 (1/2 of row 2).

(* Array: *)
Grid[Table[((2*n + 1)*9^k - 1)/2, {n, 0, 8}, {k, 0, 8}]]
(* Array antidiagonals flattened: *)
Flatten[Table[((2*(n - k) + 1)*9^k - 1)/2, {n, 0, 8}, {k, 0, n}]]

G.f. for row n: (n+(4-n)*x)/((1-x)(1-9*x)).

Recurrence for row n: A(n,k) = 10*A(n,k-1)-9*A(n,k-2), k >= 2, A(n,0) = n, A(n,1) = 9*n+4.

A333813 a(n) = 2^(1 + floor(n*log_2(3))) - (3^n + 1).

Original entry on oeis.org

0, 0, 6, 4, 46, 12, 294, 1908, 1630, 13084, 6486, 84996, 517134, 502828, 3605638, 2428308, 24062142, 5077564, 149450422, 985222180, 808182894, 6719515980, 2978678758, 43295774644, 267326277406, 252223018332, 1856180682774, 1170495537220

Offset: 0

Ctibor O. Zizka, Apr 06 2020

nonn

For integers X, Y, let a(n) = (X^(t+1) - 1) / (X - 1) - Y^n, where t = floor(n*log_X(Y)) . This sequence is for X = 2, Y = 3.

			a(0) = 2^(1 + floor(0*log_2(3))) - (3^0 + 1) = 0; a(4) = 2^(1 + floor(4*log_2(3))) - (3^4 + 1) = 46.

Cf. A000225, A024036.

Examples for integers X = Y from {2, 3, 4, 5, 6, 7, 8, 9, 10} are A000225, A003462, A002450, A003463, A003464, A023000, A023001, A002452, A002275. Examples for X = 2, Y = 4 are A024036; for X = 2, Y = 8, A024088; and for X = 3, Y = 9, A191681.

Table[2^(1+Floor[n Log2[3]])-(3^n+1),{n,0,30}] (* Harvey P. Dale, Sep 04 2023 *)

a(n) = 2^(1 + floor(n*log_2(3))) - (3^n + 1).

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A256786 Numbers which are divisible by prime(d) for all digits d in their decimal representation.

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A125857 Numbers whose base-9 representation is 22222222.......2.

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A350993 Triangular numbers that are palindromes in base 9.

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A255043 a(n) = (5*9^n - 1)/2.

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A255044 Array A read by upward antidiagonals: A(n,k) = ((2*n+1)*9^k-1)/2, n,k >= 0.

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A333813 a(n) = 2^(1 + floor(n*log_2(3))) - (3^n + 1).

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A255044 Array A read by upward antidiagonals: A(n,k) = ((2n+1)9^k-1)/2, n,k >= 0.