A037144 -id:A037144 - cp's OEIS frontend

A001399 a(n) is the number of partitions of n into at most 3 parts; also partitions of n+3 in which the greatest part is 3; also number of unlabeled multigraphs with 3 nodes and n edges.

1, 1, 2, 3, 4, 5, 7, 8, 10, 12, 14, 16, 19, 21, 24, 27, 30, 33, 37, 40, 44, 48, 52, 56, 61, 65, 70, 75, 80, 85, 91, 96, 102, 108, 114, 120, 127, 133, 140, 147, 154, 161, 169, 176, 184, 192, 200, 208, 217, 225, 234, 243, 252, 261, 271, 280, 290, 300, 310, 320, 331, 341

Offset: 0

N. J. A. Sloane

Also number of tripods (trees with exactly 3 leaves) on n vertices. - Eric W. Weisstein, Mar 05 2011
Also number of partitions of n+3 into exactly 3 parts; number of partitions of n in which the greatest part is less than or equal to 3; and the number of nonnegative solutions to b + 2c + 3d = n.
Also a(n) gives number of partitions of n+6 into 3 distinct parts and number of partitions of 2n+9 into 3 distinct and odd parts, e.g., 15 = 11 + 3 + 1 = 9 + 5 + 1 = 7 + 5 + 3. - Jon Perry, Jan 07 2004
Also bracelets with n+3 beads 3 of which are red (so there are 2 possibilities with 5 beads).
More generally, the number of partitions of n into at most k parts is also the number of partitions of n+k into k positive parts, the number of partitions of n+k in which the greatest part is k, the number of partitions of n in which the greatest part is less than or equal to k, the number of partitions of n+k(k+1)/2 into exactly k distinct positive parts, the number of nonnegative solutions to b + 2c + 3d + ... + kz = n and the number of nonnegative solutions to 2c + 3d + ... + kz <= n. - Henry Bottomley, Apr 17 2001
Also coefficient of q^n in the expansion of (m choose 3)_q as m goes to infinity. - Y. Kelly Itakura (yitkr(AT)mta.ca), Aug 21 2002
From Winston C. Yang (winston(AT)cs.wisc.edu), Apr 30 2002: (Start)
Write 1,2,3,4,... in a hexagonal spiral around 0, then a(n) for n > 0 is formed by the folding points (including the initial 1). The spiral begins:
.
                 85--84--83--82--81--80
                 /                     \
               86  56--55--54--53--52  79
               /   /                 \   \
             87  57  33--32--31--30  51  78
             /   /   /             \   \   \
           88  58  34  16--15--14  29  50  77
           /   /   /   /         \   \   \   \
         89  59  35  17   5---4  13  28  49  76
         /   /   /   /   /     \   \   \   \   \
       90  60  36  18   6   0   3  12  27  48  75
       /   /   /   /   /   /   /   /   /   /   /
     91  61  37  19   7   1---2  11  26  47  74
           \   \   \   \         /   /   /   /
           62  38  20   8---9--10  25  46  73
             \   \   \             /   /   /
             63  39  21--22--23--24  45  72
               \   \                 /   /
               64  40--41--42--43--44  71
                 \                     /
                 65--66--67--68--69--70
.
a(p) is maximal number of hexagons in a polyhex with perimeter at most 2p + 6. (End)
a(n-3) is the number of partitions of n into 3 distinct parts, where 0 is allowed as a part. E.g., at n=9, we can write 8+1+0, 7+2+0, 6+3+0, 4+5+0, 1+2+6, 1+3+5 and 2+3+4, which is a(6)=7. - Jon Perry, Jul 08 2003
a(n) gives number of partitions of n+6 into parts <=3 where each part is used at least once (subtract 6=1+2+3 from n). - Jon Perry, Jul 03 2004
This is also the number of partitions of n+3 into exactly 3 parts (there is a 1-to-1 correspondence between the number of partitions of n+3 in which the greatest part is 3 and the number of partitions of n+3 into exactly three parts). - Graeme McRae, Feb 07 2005
Apply the Riordan array (1/(1-x^3),x) to floor((n+2)/2). - Paul Barry, Apr 16 2005
Also, number of triangles that can be created with odd perimeter 3,5,7,9,11,... with all sides whole numbers. Note that triangles with even perimeter can be generated from the odd ones by increasing each side by 1. E.g., a(1) = 1 because perimeter 3 can make {1,1,1} 1 triangle. a(4) = 3 because perimeter 9 can make {1,4,4} {2,3,4} {3,3,3} 3 possible triangles. - Bruce Love (bruce_love(AT)ofs.edu.sg), Nov 20 2006
Also number of nonnegative solutions of the Diophantine equation x+2*y+3*z=n, cf. Pólya/Szegő reference.
From Vladimir Shevelev, Apr 23 2011: (Start)
Also a(n-3), n >= 3, is the number of non-equivalent necklaces of 3 beads each of them painted by one of n colors.
The sequence {a(n-3), n >= 3} solves the so-called Reis problem about convex k-gons in case k=3 (see our comment to A032279).
a(n-3) (n >= 3) is an essentially unimprovable upper estimate for the number of distinct values of the permanent in (0,1)-circulants of order n with three 1's in every row. (End)
A001399(n) is the number of 3-tuples (w,x,y) having all terms in {0,...,n} and w = 2*x+3*y. - Clark Kimberling, Jun 04 2012
Also, for n >= 3, a(n-3) is the number of the distinct triangles in an n-gon, see the Ngaokrajang links. - Kival Ngaokrajang, Mar 16 2013
Also, a(n) is the total number of 5-curve coin patterns (5C4S type: 5 curves covering full 4 coins and symmetry) packing into fountain of coins base (n+3). See illustration in links. - Kival Ngaokrajang, Oct 16 2013
Also a(n) = half the number of minimal zero sequences for Z_n of length 3 [Ponomarenko]. - N. J. A. Sloane, Feb 25 2014
Also, a(n) equals the number of linearly-independent terms at 2n-th order in the power series expansion of an Octahedral Rotational Energy Surface (cf. Harter & Patterson). - Bradley Klee, Jul 31 2015
Also Molien series for invariants of finite Coxeter groups D_3 and A_3. - N. J. A. Sloane, Jan 10 2016
Number of different distributions of n+6 identical balls in 3 boxes as x,y,z where 0 < x < y < z. - Ece Uslu and Esin Becenen, Jan 11 2016
a(n) is also the number of partitions of 2*n with <= n parts and no part >= 4. The bijection to partitions of n with no part >= 4 is: 1 <-> 2, 2 <-> 1 + 3, 3 <-> 3 + 3 (observing the order of these rules). The <- direction uses the following fact for partitions of 2*n with <= n parts and no part >=4: for each part 1 there is a part 3, and an even number (including 0) of remaining parts 3. - Wolfdieter Lang, May 21 2019
List of the terms in A000567(n>=1), A049450(n>=1), A033428(n>=1), A049451(n>=1), A045944(n>=1), and A003215(n) in nondecreasing order. List of the numbers A056105(n)-1, A056106(n)-1, A056107(n)-1, A056108(n)-1, A056109(n)-1, and A003215(m) with n >= 1 and m >= 0 in nondecreasing order. Numbers of the forms 3n*(n-1)+1, n*(3n-2), n*(3n-1), 3n^2, n*(3n+1), n*(3n+2) with n >= 1 listed in nondecreasing order. Integers m such that lattice points from 1 through m on a hexagonal spiral starting at 1 forms a convex polygon. - Ya-Ping Lu, Jan 24 2024

			G.f. = 1 + x + 2*x^2 + 3*x^3 + 4*x^4 + 5*x^5 + 7*x^6 + 8*x^7 + 10*x^8 + 12*x^9 + ...
Recall that in a necklace the adjacent beads have distinct colors. Suppose we have n colors with labels 1,...,n. Two colorings of the beads are equivalent if the cyclic sequences of the distances modulo n between labels of adjacent colors have the same period. If n=4, all colorings are equivalent. E.g., for the colorings {1,2,3} and {1,2,4} we have the same period {1,1,2} of distances modulo 4. So, a(n-3)=a(1)=1. If n=5, then we have two such periods {1,1,3} and {1,2,2} modulo 5. Thus a(2)=2. - _Vladimir Shevelev_, Apr 23 2011
a(0) = 1, i.e., {1,2,3} Number of different distributions of 6 identical balls to 3 boxes as x,y and z where 0 < x < y < z. - _Ece Uslu_, Esin Becenen, Jan 11 2016
a(3) = 3, i.e., {1,2,6}, {1,3,5}, {2,3,4} Number of different distributions of 9 identical balls in 3 boxes as x,y and z where 0 < x < y < z. - _Ece Uslu_, Esin Becenen, Jan 11 2016
From _Gus Wiseman_, Apr 15 2019: (Start)
The a(0) = 1 through a(8) = 10 integer partitions of n with at most three parts are the following. The Heinz numbers of these partitions are given by A037144.
  ()  (1)  (2)   (3)    (4)    (5)    (6)    (7)    (8)
           (11)  (21)   (22)   (32)   (33)   (43)   (44)
                 (111)  (31)   (41)   (42)   (52)   (53)
                        (211)  (221)  (51)   (61)   (62)
                               (311)  (222)  (322)  (71)
                                      (321)  (331)  (332)
                                      (411)  (421)  (422)
                                             (511)  (431)
                                                    (521)
                                                    (611)
The a(0) = 1 through a(7) = 8 integer partitions of n + 3 whose greatest part is 3 are the following. The Heinz numbers of these partitions are given by A080193.
  (3)  (31)  (32)   (33)    (322)    (332)     (333)      (3322)
             (311)  (321)   (331)    (3221)    (3222)     (3331)
                    (3111)  (3211)   (3311)    (3321)     (32221)
                            (31111)  (32111)   (32211)    (33211)
                                     (311111)  (33111)    (322111)
                                               (321111)   (331111)
                                               (3111111)  (3211111)
                                                          (31111111)
Non-isomorphic representatives of the a(0) = 1 through a(5) = 5 unlabeled multigraphs with 3 vertices and n edges are the following.
  {}  {12}  {12,12}  {12,12,12}  {12,12,12,12}  {12,12,12,12,12}
            {13,23}  {12,13,23}  {12,13,23,23}  {12,13,13,23,23}
                     {13,23,23}  {13,13,23,23}  {12,13,23,23,23}
                                 {13,23,23,23}  {13,13,23,23,23}
                                                {13,23,23,23,23}
The a(0) = 1 through a(8) = 10 strict integer partitions of n - 6 with three parts are the following (A = 10, B = 11). The Heinz numbers of these partitions are given by A007304.
  (321)  (421)  (431)  (432)  (532)  (542)  (543)  (643)   (653)
                (521)  (531)  (541)  (632)  (642)  (652)   (743)
                       (621)  (631)  (641)  (651)  (742)   (752)
                              (721)  (731)  (732)  (751)   (761)
                                     (821)  (741)  (832)   (842)
                                            (831)  (841)   (851)
                                            (921)  (931)   (932)
                                                   (A21)   (941)
                                                           (A31)
                                                           (B21)
The a(0) = 1 through a(8) = 10 integer partitions of n + 3 with three parts are the following. The Heinz numbers of these partitions are given by A014612.
  (111)  (211)  (221)  (222)  (322)  (332)  (333)  (433)  (443)
                (311)  (321)  (331)  (422)  (432)  (442)  (533)
                       (411)  (421)  (431)  (441)  (532)  (542)
                              (511)  (521)  (522)  (541)  (551)
                                     (611)  (531)  (622)  (632)
                                            (621)  (631)  (641)
                                            (711)  (721)  (722)
                                                   (811)  (731)
                                                          (821)
                                                          (911)
The a(0) = 1 through a(8) = 10 integer partitions of n whose greatest part is <= 3 are the following. The Heinz numbers of these partitions are given by A051037.
  ()  (1)  (2)   (3)    (22)    (32)     (33)      (322)      (332)
           (11)  (21)   (31)    (221)    (222)     (331)      (2222)
                 (111)  (211)   (311)    (321)     (2221)     (3221)
                        (1111)  (2111)   (2211)    (3211)     (3311)
                                (11111)  (3111)    (22111)    (22211)
                                         (21111)   (31111)    (32111)
                                         (111111)  (211111)   (221111)
                                                   (1111111)  (311111)
                                                              (2111111)
                                                              (11111111)
The a(0) = 1 through a(6) = 7 strict integer partitions of 2n+9 with 3 parts, all of which are odd, are the following. The Heinz numbers of these partitions are given by A307534.
  (5,3,1)  (7,3,1)  (7,5,1)  (7,5,3)   (9,5,3)   (9,7,3)   (9,7,5)
                    (9,3,1)  (9,5,1)   (9,7,1)   (11,5,3)  (11,7,3)
                             (11,3,1)  (11,5,1)  (11,7,1)  (11,9,1)
                                       (13,3,1)  (13,5,1)  (13,5,3)
                                                 (15,3,1)  (13,7,1)
                                                           (15,5,1)
                                                           (17,3,1)
The a(0) = 1 through a(8) = 10 strict integer partitions of n + 3 with 3 parts where 0 is allowed as a part (A = 10):
  (210)  (310)  (320)  (420)  (430)  (530)  (540)  (640)  (650)
                (410)  (510)  (520)  (620)  (630)  (730)  (740)
                       (321)  (610)  (710)  (720)  (820)  (830)
                              (421)  (431)  (810)  (910)  (920)
                                     (521)  (432)  (532)  (A10)
                                            (531)  (541)  (542)
                                            (621)  (631)  (632)
                                                   (721)  (641)
                                                          (731)
                                                          (821)
The a(0) = 1 through a(7) = 7 integer partitions of n + 6 whose distinct parts are 1, 2, and 3 are the following. The Heinz numbers of these partitions are given by A143207.
  (321)  (3211)  (3221)   (3321)    (32221)    (33221)     (33321)
                 (32111)  (32211)   (33211)    (322211)    (322221)
                          (321111)  (322111)   (332111)    (332211)
                                    (3211111)  (3221111)   (3222111)
                                               (32111111)  (3321111)
                                                           (32211111)
                                                           (321111111)
(End)
Partitions of 2*n with <= n parts and no part >= 4: a(3) = 3 from (2^3), (1,2,3), (3^2) mapping to (1^3), (1,2), (3), the partitions of 3 with no part >= 4, respectively. - _Wolfdieter Lang_, May 21 2019

R. Ayoub, An Introduction to the Analytic Theory of Numbers, Amer. Math. Soc., 1963; Chapter III, Problem 33.
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N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
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Marius A. Burtea, Table of n, a(n) for n = 0..17501 (terms 0..1000 from T. D. Noe, terms 14001 onwards corrected by Sean A. Irvine, April 25 2019)
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Index entries for linear recurrences with constant coefficients, signature (1,1,0,-1,-1,1).
Index entries for Molien series.
Index entries for two-way infinite sequences

Cf. A008724, A003082, A117485, A026810, A026811, A026812, A026813, A026814, A026815, A026816, A000228, A036496, A008619, A001400, A001401, A069905, A008615, row 3 of A192517.
Molien series for finite Coxeter groups D_3 through D_12 are A001399, A051263, A266744-A266751.
Cf. A007304, A014612, A037144, A051037, A080193, A143207, A307534.

a001399 = p [1,2,3] where
   p _      0 = 1
   p []     _ = 0
   p ks'@(k:ks) m = if m < k then 0 else p ks' (m - k) + p ks m
-- Reinhard Zumkeller, Feb 28 2013

I:=[1,1,2,3,4,5]; [n le 6 select I[n] else Self(n-1)+Self(n-2)-Self(n-4)-Self(n-5)+Self(n-6): n in [1..80]]; // Vincenzo Librandi, Feb 14 2015

[#RestrictedPartitions(n,{1,2,3}): n in [0..62]]; // Marius A. Burtea, Jan 06 2019

[Round((n+3)^2/12): n in [0..70]]; // Marius A. Burtea, Jan 06 2019

A001399 := proc(n)
    round( (n+3)^2/12) ;
end proc:
seq(A001399(n),n=0..40) ;
with(combstruct):ZL4:=[S,{S=Set(Cycle(Z,card<4))}, unlabeled]:seq(count(ZL4,size=n),n=0..61); # Zerinvary Lajos, Sep 24 2007
B:=[S,{S = Set(Sequence(Z,1 <= card),card <=3)},unlabelled]: seq(combstruct[count](B, size=n), n=0..61); # Zerinvary Lajos, Mar 21 2009

CoefficientList[ Series[ 1/((1 - x)*(1 - x^2)*(1 - x^3)), {x, 0, 65} ], x ]
Table[ Length[ IntegerPartitions[n, 3]], {n, 0, 61} ] (* corrected by Jean-François Alcover, Aug 08 2012 *)
k = 3; Table[(Apply[Plus, Map[EulerPhi[ # ]Binomial[n/#, k/# ] &, Divisors[GCD[n, k]]]]/n + Binomial[If[OddQ[n], n - 1, n - If[OddQ[k], 2, 0]]/2, If[OddQ[k], k - 1, k]/2])/2, {n, k, 50}] (* Robert A. Russell, Sep 27 2004 *)
LinearRecurrence[{1,1,0,-1,-1,1},{1,1,2,3,4,5},70] (* Harvey P. Dale, Jun 21 2012 *)
a[ n_] := With[{m = Abs[n + 3] - 3}, Length[ IntegerPartitions[ m, 3]]]; (* Michael Somos, Dec 25 2014 *)
k=3 (* Number of red beads in bracelet problem *);CoefficientList[Series[(1/k Plus@@(EulerPhi[#] (1-x^#)^(-(k/#))&/@Divisors[k])+(1+x)/(1-x^2)^Floor[(k+2)/2])/2,{x,0,50}],x] (* Herbert Kociemba, Nov 04 2016 *)
Table[Length[Select[IntegerPartitions[n,{3}],UnsameQ@@#&]],{n,0,30}] (* Gus Wiseman, Apr 15 2019 *)

{a(n) = round((n + 3)^2 / 12)}; /* Michael Somos, Sep 04 2006 */

[round((n+3)**2 / 12) for n in range(0,62)] # Ya-Ping Lu, Jan 24 2024

G.f.: 1/((1 - x) * (1 - x^2) * (1 - x^3)) = -1/((x+1)*(x^2+x+1)*(x-1)^3); Simon Plouffe in his 1992 dissertation
a(n) = round((n + 3)^2/12). Note that this cannot be of the form (2*i + 1)/2, so ties never arise.
a(n) = A008284(n+3, 3), n >= 0.
a(n) = 1 + a(n-2) + a(n-3) - a(n-5) for all n in Z. - Michael Somos, Sep 04 2006
a(n) = a(-6 - n) for all n in Z. - Michael Somos, Sep 04 2006
a(6*n) = A003215(n), a(6*n + 1) = A000567(n + 1), a(6*n + 2) = A049450(n + 1), a(6*n + 3) = A033428(n + 1), a(6*n + 4) = A049451(n + 1), a(6*n + 5) = A045944(n + 1).
a(n) = a(n-1) + A008615(n+2) = a(n-2) + A008620(n) = a(n-3) + A008619(n) = A001840(n+1) - a(n-1) = A002620(n+2) - A001840(n) = A000601(n) - A000601(n-1). - Henry Bottomley, Apr 17 2001
P(n, 3) = (1/72) * (6*n^2 - 7 - 9*pcr{1, -1}(2, n) + 8*pcr{2, -1, -1}(3, n)) (see Comtet). [Here "pcr" stands for "prime circulator" and it is defined on p. 109 of Comtet, while the formula appears on p. 110. - Petros Hadjicostas, Oct 03 2019]
Let m > 0 and -3 <= p <= 2 be defined by n = 6*m+p-3; then for n > -3, a(n) = 3*m^2 + p*m, and for n = -3, a(n) = 3*m^2 + p*m + 1. - Floor van Lamoen, Jul 23 2001
72*a(n) = 17 + 6*(n+1)*(n+5) + 9*(-1)^n - 8*A061347(n). - Benoit Cloitre, Feb 09 2003
From Jon Perry, Jun 17 2003: (Start)
a(n) = 6*t(floor(n/6)) + (n%6) * (floor(n/6) + 1) + (n mod 6 == 0?1:0), where t(n) = n*(n+1)/2.
a(n) = ceiling(1/12*n^2 + 1/2*n) + (n mod 6 == 0?1:0).
[Here "n%6" means "n mod 6" while "(n mod 6 == 0?1:0)" means "if n mod 6 == 0 then 1, else 0" (as in C).]
(End)
a(n) = Sum_{i=0..floor(n/3)} 1 + floor((n - 3*i)/2). - Jon Perry, Jun 27 2003
a(n) = Sum_{k=0..n} floor((k + 2)/2) * (cos(2*Pi*(n - k)/3 + Pi/3)/3 + sqrt(3) * sin(2*Pi*(n-k)/3 + Pi/3)/3 + 1/3). - Paul Barry, Apr 16 2005
(m choose 3)_q = (q^m-1) * (q^(m-1) - 1) * (q^(m-2) - 1)/((q^3 - 1) * (q^2 - 1) * (q - 1)).
a(n) = Sum_{k=0..floor(n/2)} floor((3 + n - 2*k)/3). - Paul Barry, Nov 11 2003
A117220(n) = a(A003586(n)). - Reinhard Zumkeller, Mar 04 2006
a(n) = 3 * Sum_{i=2..n+1} floor(i/2) - floor(i/3). - Thomas Wieder, Feb 11 2007
Identical to the number of points inside or on the boundary of the integer grid of {I, J}, bounded by the three straight lines I = 0, I - J = 0 and I + 2J = n. - Jonathan Vos Post, Jul 03 2007
a(n) = A026820(n,3) for n > 2. - Reinhard Zumkeller, Jan 21 2010
Euler transform of length 3 sequence [ 1, 1, 1]. - Michael Somos, Feb 25 2012
a(n) = A005044(2*n + 3) = A005044(2*n + 6). - Michael Somos, Feb 25 2012
a(n) = A000212(n+3) - A002620(n+3). - Richard R. Forberg, Dec 08 2013
a(n) = a(n-1) + a(n-2) - a(n-4) - a(n-5) + a(n-6). - David Neil McGrath, Feb 14 2015
a(n) = floor((n^2+3)/12) + floor((n+2)/2). - Giacomo Guglieri, Apr 02 2019
From Devansh Singh, May 28 2020: (Start)
Let p(n, 3) be the number of 3-part integer partitions in which every part is > 0.
Then for n >= 3, p(n, 3) is equal to:
  (n^2 - 1)/12 when n is odd and 3 does not divide n.
  (n^2 + 3)/12 when n is odd and 3 divides n.
  (n^2 - 4)/12 when n is even and 3 does not divide n.
  (n^2)/12 when n is even and 3 divides n.
For n >= 3, p(n, 3) = a(n-3). (End)
a(n) = floor(((n+3)^2 + 4)/12). - Vladimír Modrák, Zuzana Soltysova, Dec 08 2020
Sum_{n>=0} 1/a(n) = 15/4 - Pi/(2*sqrt(3)) + Pi^2/18 + tanh(Pi/(2*sqrt(3)))*Pi/sqrt(3). - Amiram Eldar, Sep 29 2022
E.g.f.: exp(-x)*(9 + exp(2*x)*(47 + 42*x + 6*x^2) + 16*exp(x/2)*cos(sqrt(3)*x/2))/72. - Stefano Spezia, Mar 05 2023
a(6n) = 1+6*A000217(n); Sum_{i=1..n} a(6*i) = A000578(n+1). - David García Herrero, May 05 2024

Name edited by Gus Wiseman, Apr 15 2019

A051037 5-smooth numbers, i.e., numbers whose prime divisors are all <= 5.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30, 32, 36, 40, 45, 48, 50, 54, 60, 64, 72, 75, 80, 81, 90, 96, 100, 108, 120, 125, 128, 135, 144, 150, 160, 162, 180, 192, 200, 216, 225, 240, 243, 250, 256, 270, 288, 300, 320, 324, 360, 375, 384, 400, 405

Offset: 1

Eric W. Weisstein

Sometimes called the Hamming sequence, since Hamming asked for an efficient algorithm to generate the list, in ascending order, of all numbers of the form 2^i*3^j*5^k for i,j,k >= 0. The problem was popularized by Edsger Dijkstra.
Numbers k such that 8*k = EulerPhi(30*k). - Artur Jasinski, Nov 05 2008
Where record values greater than 1 occur in A165704: A165705(n) = A165704(a(n)). - Reinhard Zumkeller, Sep 26 2009
Also called "harmonic whole numbers", see Howard and Longair, 1982, Table I, page 121. - Hugo Pfoertner, Jul 16 2020
Also called ugly numbers, although it is not clear why. - Gus Wiseman, May 21 2021
Some woody bamboo species have extraordinarily long and stable flowering intervals that belong to this sequence. The model by Veller, Nowak & Davis justifies this observation from the evolutionary point of view. - Andrey Zabolotskiy, Jun 27 2021
Also those integers k for which, for every prime p > 5, p^(4*k) - 1 == 0 (mod 240*k). - Federico Provvedi, May 23 2022
As noted in the comments to A085152, Størmer's theorem implies that the only pairs of consecutive integers that appear as consecutive terms of this sequence are (1,2), (2,3), (3,4), (4,5), (5,6), (8,9), (9,10), (15,16), (24,25), and (80,81). These all represent significant musical intervals. - Hal M. Switkay, Dec 05 2022

			From _Gus Wiseman_, May 21 2021: (Start)
The sequence of terms together with their prime indices begins:
      1: {}            25: {3,3}
      2: {1}           27: {2,2,2}
      3: {2}           30: {1,2,3}
      4: {1,1}         32: {1,1,1,1,1}
      5: {3}           36: {1,1,2,2}
      6: {1,2}         40: {1,1,1,3}
      8: {1,1,1}       45: {2,2,3}
      9: {2,2}         48: {1,1,1,1,2}
     10: {1,3}         50: {1,3,3}
     12: {1,1,2}       54: {1,2,2,2}
     15: {2,3}         60: {1,1,2,3}
     16: {1,1,1,1}     64: {1,1,1,1,1,1}
     18: {1,2,2}       72: {1,1,1,2,2}
     20: {1,1,3}       75: {2,3,3}
     24: {1,1,1,2}     80: {1,1,1,1,3}
(End)

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000 (first 1000 terms from T. D. Noe)
Benoit Cloitre, Plot of abs(f(n)-s(n)) vs its mean values (blue) and vs loglog(n) (red).
M. J. Dominus, Infinite Lists in Perl.
Deborah Howard and Malcolm Longair, Harmonic Proportion and Palladio's "Quattro Libri", Journal of the Society of Architectural Historians (1982) 41 (2): 116-143.
Vaclav Kotesovec, Plot of a(n) / (exp((6*log(2)*log(3)*log(5)*n)^(1/3))/sqrt(30)) for n = 1..1200000
Rosetta Code, A collection of computer codes to compute 5-smooth numbers.
Raphael Schumacher, The Formula for the Distribution of the 3-Smooth Numbers, 5-Smooth, 7-Smooth and all other Smooth Numbers, arXiv:1608.06928 [math.NT], 2016.
Sci.math, Ugly numbers.
Carl Veller, Martin A. Nowak and Charles C. Davis, Extended flowering intervals of bamboos evolved by discrete multiplication, Ecology Letters, 18 (2015), 653-659.
Eric Weisstein's World of Mathematics, Smooth Number.
Wikipedia, Regular number.
Wikipedia, Talk:Regular number. Includes a discussion of the name.
Wikipedia, Størmer's theorem.

Subsequences: A003592, A003593, A051916 , A257997.
For p-smooth numbers with other values of p, see A003586, A002473, A051038, A080197, A080681, A080682, A080683.
Cf. A006530, A112757, A112758, A112759, A112763, A112764, A291719.
The partitions with these Heinz numbers are counted by A001399.
The conjugate opposite is A033942, counted by A004250.
The opposite is A059485, counted by A004250.
The non-3-smooth case is A080193, counted by A069905.
The conjugate is A037144, counted by A001399.
The complement is A279622, counted by A035300.
Requiring the sum of prime indices to be even gives A344297.
Cf. A000244, A002182, A002183, A035301, A056239, A112798, A261144, A344293.

import Data.Set (singleton, deleteFindMin, insert)
a051037 n = a051037_list !! (n-1)
a051037_list = f $ singleton 1 where
   f s = y : f (insert (5 * y) $ insert (3 * y) $ insert (2 * y) s')
               where (y, s') = deleteFindMin s
-- Reinhard Zumkeller, May 16 2015

[n: n in [1..500] | PrimeDivisors(n) subset [2,3,5]]; // Bruno Berselli, Sep 24 2012

A051037 := proc(n)
    option remember;
    local a;
    if n = 1 then
        1;
    else
        for a from procname(n-1)+1 do
            numtheory[factorset](a) minus {2, 3,5 } ;
            if % = {} then
                return a;
            end if;
        end do:
    end if;
end proc:
seq(A051037(n),n=1..100) ; # R. J. Mathar, Nov 05 2017

mx = 405; Sort@ Flatten@ Table[ 2^a*3^b*5^c, {a, 0, Log[2, mx]}, {b, 0, Log[3, mx/2^a]}, {c, 0, Log[5, mx/(2^a*3^b)]}] (* Or *)
Select[ Range@ 405, Last@ Map[First, FactorInteger@ #] < 7 &] (* Robert G. Wilson v *)
With[{nn=10},Select[Union[Times@@@Flatten[Table[Tuples[{2,3,5},n],{n,0,nn}],1]],#<=2^nn&]] (* Harvey P. Dale, Feb 28 2022 *)

test(n)= {m=n; forprime(p=2,5, while(m%p==0,m=m/p)); return(m==1)}
for(n=1,500,if(test(n),print1(n",")))

a(n)=local(m); if(n<1,0,n=a(n-1); until(if(m=n, forprime(p=2,5, while(m%p==0,m/=p)); m==1),n++); n)

list(lim)=my(v=List(),s,t); for(i=0,logint(lim\=1,5), t=5^i; for(j=0,logint(lim\t,3), s=t*3^j; while(s<=lim, listput(v,s); s<<=1))); Set(v) \\ Charles R Greathouse IV, Sep 21 2011; updated Sep 19 2016

smooth(P:vec,lim)={ my(v=List([1]),nxt=vector(#P,i,1),indx,t);
while(1, t=vecmin(vector(#P,i,v[nxt[i]]*P[i]),&indx);
if(t>lim,break); if(t>v[#v],listput(v,t)); nxt[indx]++);
Vec(v)
};
smooth([2,3,5], 1e4) \\ Charles R Greathouse IV, Dec 03 2013

is_A051037(n)=n<7||vecmax(factor(n,6)[, 1])<7 \\ M. F. Hasler, Jan 16 2015

def isok(n):
  while n & 1 == 0: n >>= 1
  while n % 3 == 0: n //= 3
  while n % 5 == 0: n //= 5
  return n == 1 #  Darío Clavijo, Dec 30 2022

from sympy import integer_log
def A051037(n):
    def bisection(f,kmin=0,kmax=1):
        while f(kmax) > kmax: kmax <<= 1
        while kmax-kmin > 1:
            kmid = kmax+kmin>>1
            if f(kmid) <= kmid:
                kmax = kmid
            else:
                kmin = kmid
        return kmax
    def f(x):
        c = n+x
        for i in range(integer_log(x,5)[0]+1):
            for j in range(integer_log(y:=x//5**i,3)[0]+1):
                c -= (y//3**j).bit_length()
        return c
    return bisection(f,n,n) # Chai Wah Wu, Sep 16 2024

# faster for initial segment of sequence
import heapq
from itertools import islice
def A051037gen(): # generator of terms
    v, oldv, h, psmooth_primes, = 1, 0, [1], [2, 3, 5]
    while True:
        v = heapq.heappop(h)
        if v != oldv:
            yield v
            oldv = v
            for p in psmooth_primes:
                    heapq.heappush(h, v*p)
print(list(islice(A051037gen(), 65))) # Michael S. Branicky, Sep 17 2024

Let s(n) = Card(k | a(k)Benoit Cloitre, Dec 30 2001
The characteristic function of this sequence is given by:
  Sum_{n>=1} x^a(n) = Sum_{n>=1} -Möbius(30*n)*x^n/(1-x^n). - Paul D. Hanna, Sep 18 2011
a(n) = A143207(n) / 30. - Reinhard Zumkeller, Sep 13 2011
A204455(15*a(n)) = 15, and only for these numbers. - Wolfdieter Lang, Feb 04 2012
A006530(a(n)) <= 5. - Reinhard Zumkeller, May 16 2015
Sum_{n>=1} 1/a(n) = Product_{primes p <= 5} p/(p-1) = (2*3*5)/(1*2*4) = 15/4. - Amiram Eldar, Sep 22 2020

A101296 n has the a(n)-th distinct prime signature.

Original entry on oeis.org

1, 2, 2, 3, 2, 4, 2, 5, 3, 4, 2, 6, 2, 4, 4, 7, 2, 6, 2, 6, 4, 4, 2, 8, 3, 4, 5, 6, 2, 9, 2, 10, 4, 4, 4, 11, 2, 4, 4, 8, 2, 9, 2, 6, 6, 4, 2, 12, 3, 6, 4, 6, 2, 8, 4, 8, 4, 4, 2, 13, 2, 4, 6, 14, 4, 9, 2, 6, 4, 9, 2, 15, 2, 4, 6, 6, 4, 9, 2, 12, 7, 4, 2, 13, 4, 4, 4, 8, 2, 13, 4, 6, 4, 4, 4, 16, 2, 6, 6, 11, 2, 9, 2, 8, 9, 4, 2, 15, 2, 9, 4, 12, 2, 9, 4, 6, 6, 4, 4, 17

Offset: 1

David Wasserman, Dec 21 2004

From Antti Karttunen, May 12 2017: (Start)
Restricted growth sequence transform of A046523, the least representative of each prime signature. Thus this partitions the natural numbers to the same equivalence classes as A046523, i.e., for all i, j: a(i) = a(j) <=> A046523(i) = A046523(j), and for that reason satisfies in that respect all the same conditions as A046523. For example, we have, for all i, j: if a(i) = a(j), then:
    A000005(i) = A000005(j), A008683(i) = A008683(j), A286605(i) = A286605(j).
So, this sequence (instead of A046523) can be used for finding sequences where a(n)'s value is dependent only on the prime signature of n, that is, only on the multiset of prime exponents in the factorization of n. (End)
This is also the restricted growth sequence transform of many other sequences, for example, that of A181819. See further comments there. - Antti Karttunen, Apr 30 2022

			From _David A. Corneth_, May 12 2017: (Start)
1 has prime signature (), the first distinct prime signature. Therefore, a(1) = 1.
2 has prime signature (1), the second distinct prime signature after (1). Therefore, a(2) = 2.
3 has prime signature (1), as does 2. Therefore, a(3) = a(2) = 2.
4 has prime signature (2), the third distinct prime signature after () and (1). Therefore, a(4) = 3. (End)
From _Antti Karttunen_, May 12 2017: (Start)
Construction of restricted growth sequences: In this case we start with a(1) = 1 for A046523(1) = 1, and thereafter, for all n > 1, we use the least so far unused natural number k for a(n) if A046523(n) has not been encountered before, otherwise [whenever A046523(n) = A046523(m), for some m < n], we set a(n) = a(m).
For n = 2, A046523(2) = 2, which has not been encountered before (first prime), thus we allot for a(2) the least so far unused number, which is 2, thus a(2) = 2.
For n = 3, A046523(2) = 2, which was already encountered as A046523(1), thus we set a(3) = a(2) = 2.
For n = 4, A046523(4) = 4, not encountered before (first square of prime), thus we allot for a(4) the least so far unused number, which is 3, thus a(4) = 3.
For n = 5, A046523(5) = 2, as for the first time encountered at n = 2, thus we set a(5) = a(2) = 2.
For n = 6, A046523(6) = 6, not encountered before (first semiprime pq with distinct p and q), thus we allot for a(6) the least so far unused number, which is 4, thus a(6) = 4.
For n = 8, A046523(8) = 8, not encountered before (first cube of a prime), thus we allot for a(8) the least so far unused number, which is 5, thus a(8) = 5.
For n = 9, A046523(9) = 4, as for the first time encountered at n = 4, thus a(9) = 3.
(End)
From _David A. Corneth_, May 12 2017: (Start)
(Rough) description of an algorithm of computing the sequence:
Suppose we want to compute a(n) for n in [1..20].
We set up a vector of 20 elements, values 0, and a number m = 1, the minimum number we haven't checked and c = 0, the number of distinct prime signatures we've found so far.
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
We check the prime signature of m and see that it's (). We increase c with 1 and set all elements up to 20 with prime signature () to 1. In the process, we adjust m. This gives:
[1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]. The least number we haven't checked is m = 2. 2 has prime signature (1). We increase c with 1 and set all elements up to 20 with prime signature (1) to 2. In the process, we adjust m. This gives:
[1, 2, 2, 0, 2, 0, 2, 0, 0, 0, 2, 0, 2, 0, 0, 0, 2, 0, 2, 0]
We check the prime signature of m = 4 and see that its prime signature is (2). We increase c with 1 and set all numbers up to 20 with prime signature (2) to 3. This gives:
[1, 2, 2, 3, 2, 0, 2, 0, 3, 0, 2, 0, 2, 0, 0, 0, 2, 0, 2, 0]
Similarily, after m = 6, we get
[1, 2, 2, 3, 2, 4, 2, 0, 3, 4, 2, 0, 2, 4, 4, 0, 2, 0, 2, 0], after m = 8 we get:
[1, 2, 2, 3, 2, 4, 2, 5, 3, 4, 2, 0, 2, 4, 4, 0, 2, 0, 2, 0], after m = 12 we get:
[1, 2, 2, 3, 2, 4, 2, 5, 3, 4, 2, 6, 2, 4, 4, 0, 2, 6, 2, 0], after m = 16 we get:
[1, 2, 2, 3, 2, 4, 2, 5, 3, 4, 2, 6, 2, 4, 4, 7, 2, 6, 2, 0], after m = 20 we get:
[1, 2, 2, 3, 2, 4, 2, 5, 3, 4, 2, 6, 2, 4, 4, 7, 2, 6, 2, 8]. Now, m > 20 so we stop. (End)
The above method is inefficient, because the step "set all elements a(n) up to n = Nmax with prime signature s(n) = S[c] to c" requires factoring all integers up to Nmax (or at least comparing their signature, once computed, with S[c]) again and again. It is much more efficient to run only once over each m = 1..Nmax, compute its prime signature s(m), add it to an ordered list in case it did not occur earlier, together with its "rank" (= new size of the list), and assign that rank to a(m). The list of prime signatures is much shorter than [1..Nmax]. One can also use m'(m) := the smallest n with the prime signature of m (which is faster to compute than to search for the signature) as representative for s(m), and set a(m) := a(m'(m)). Then it is sufficient to have just one counter (number of prime signatures seen so far) as auxiliary variable, in addition to the sequence to be computed. - _M. F. Hasler_, Jul 18 2019

Michel Marcus (terms 1..10000) & Antti Karttunen, Table of n, a(n) for n = 1..100000

Cf. A025487, A046523, A064839 (ordinal transform of this sequence), A181819, and arrays A095904, A179216.
Cf. A000005, A008683.
Sequences that are unions of finite number (>= 2) of equivalence classes determined by the values that this sequence obtains (i.e., sequences mentioned in David A. Corneth's May 12 2017 formula): A001358 (A001248 U A006881, values 3 & 4), A007422 (values 1, 4, 5), A007964 (2, 3, 4, 5), A014612 (5, 6, 9), A030513 (4, 5), A037143 (1, 2, 3, 4), A037144 (1, 2, 3, 4, 5, 6, 9), A080258 (6, 7), A084116 (2, 4, 5), A167171 (2, 4), A217856 (6, 9).
Cf. also A077462, A305897 (stricter variants, with finer partitioning) and A254524, A286603, A286605, A286610, A286619, A286621, A286622, A286626, A286378 for other similarly constructed sequences.

A101296 := proc(n)
    local a046523, a;
    a046523 := A046523(n) ;
    for a from 1 do
        if A025487(a) = a046523 then
            return a;
        elif A025487(a) > a046523 then
            return -1 ;
        end if;
    end do:
end proc: # R. J. Mathar, May 26 2017

With[{nn = 120}, Function[s, Table[Position[Keys@s, k_ /; MemberQ[k, n]][[1, 1]], {n, nn}]]@ Map[#1 -> #2 & @@ # &, Transpose@ {Values@ #, Keys@ #}] &@ PositionIndex@ Table[Times @@ MapIndexed[Prime[First@ #2]^#1 &, Sort[FactorInteger[n][[All, -1]], Greater]] - Boole[n == 1], {n, nn}] ] (* Michael De Vlieger, May 12 2017, Version 10 *)

find(ps, vps) = {for (k=1, #vps, if (vps[k] == ps, return(k)););}
lisps(nn) = {vps = []; for (n=1, nn, ps = vecsort(factor(n)[,2]); ips = find(ps, vps); if (! ips, vps = concat(vps, ps); ips = #vps); print1(ips, ", "););} \\ Michel Marcus, Nov 15 2015; edited by M. F. Hasler, Jul 16 2019

rgs_transform(invec) = { my(occurrences = Map(), outvec = vector(length(invec)), u=1); for(i=1, length(invec), if(mapisdefined(occurrences,invec[i]), my(pp = mapget(occurrences, invec[i])); outvec[i] = outvec[pp] , mapput(occurrences,invec[i],i); outvec[i] = u; u++ )); outvec; };
write_to_bfile(start_offset,vec,bfilename) = { for(n=1, length(vec), write(bfilename, (n+start_offset)-1, " ", vec[n])); }
write_to_bfile(1,rgs_transform(vector(100000,n,A046523(n))),"b101296.txt");
\\ Antti Karttunen, May 12 2017

A025487(a(n)) = A046523(n).
Indices of records give A025487. - Michel Marcus, Nov 16 2015
From David A. Corneth, May 12 2017: (Start) [Corresponding characteristic function in brackets]
a(A000012(n)) = 1 (sig.: ()).      [A063524]
a(A000040(n)) = 2 (sig.: (1)).     [A010051]
a(A001248(n)) = 3 (sig.: (2)).     [A302048]
a(A006881(n)) = 4 (sig.: (1,1)).   [A280710]
a(A030078(n)) = 5 (sig.: (3)).
a(A054753(n)) = 6 (sig.: (1,2)).   [A353472]
a(A030514(n)) = 7 (sig.: (4)).
a(A065036(n)) = 8 (sig.: (1,3)).
a(A007304(n)) = 9 (sig.: (1,1,1)). [A354926]
a(A050997(n)) = 10 (sig.: (5)).
a(A085986(n)) = 11 (sig.: (2,2)).
a(A178739(n)) = 12 (sig.: (1,4)).
a(A085987(n)) = 13 (sig.: (1,1,2)).
a(A030516(n)) = 14 (sig.: (6)).
a(A143610(n)) = 15 (sig.: (2,3)).
a(A178740(n)) = 16 (sig.: (1,5)).
a(A189975(n)) = 17 (sig.: (1,1,3)).
a(A092759(n)) = 18 (sig.: (7)).
a(A189988(n)) = 19 (sig.: (2,4)).
a(A179643(n)) = 20 (sig.: (1,2,2)).
a(A189987(n)) = 21 (sig.: (1,6)).
a(A046386(n)) = 22 (sig.: (1,1,1,1)).
a(A162142(n)) = 23 (sig.: (2,2,2)).
a(A179644(n)) = 24 (sig.: (1,1,4)).
a(A179645(n)) = 25 (sig.: (8)).
a(A179646(n)) = 26 (sig.: (2,5)).
a(A163569(n)) = 27 (sig.: (1,2,3)).
a(A179664(n)) = 28 (sig.: (1,7)).
a(A189982(n)) = 29 (sig.: (1,1,1,2)).
a(A179666(n)) = 30 (sig.: (3,4)).
a(A179667(n)) = 31 (sig.: (1,1,5)).
a(A179665(n)) = 32 (sig.: (9)).
a(A189990(n)) = 33 (sig.: (2,6)).
a(A179669(n)) = 34 (sig.: (1,2,4)).
a(A179668(n)) = 35 (sig.: (1,8)).
a(A179670(n)) = 36 (sig.: (1,1,1,3)).
a(A179671(n)) = 37 (sig.: (3,5)).
a(A162143(n)) = 38 (sig.: (2,2,2)).
a(A179672(n)) = 39 (sig.: (1,1,6)).
a(A030629(n)) = 40 (sig.: (10)).
a(A179688(n)) = 41 (sig.: (1,3,3)).
a(A179689(n)) = 42 (sig.: (2,7)).
a(A179690(n)) = 43 (sig.: (1,1,2,2)).
a(A189991(n)) = 44 (sig.: (4,4)).
a(A179691(n)) = 45 (sig.: (1,2,5)).
a(A179692(n)) = 46 (sig.: (1,9)).
a(A179693(n)) = 47 (sig.: (1,1,1,4)).
a(A179694(n)) = 48 (sig.: (3,6)).
a(A179695(n)) = 49 (sig.: (2,2,3)).
a(A179696(n)) = 50 (sig.: (1,1,7)).
(End)

Data section extended to 120 terms by Antti Karttunen, May 12 2017

Minor edits/corrections by M. F. Hasler, Jul 18 2019

A037143 Numbers with at most 2 prime factors (counted with multiplicity).

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 9, 10, 11, 13, 14, 15, 17, 19, 21, 22, 23, 25, 26, 29, 31, 33, 34, 35, 37, 38, 39, 41, 43, 46, 47, 49, 51, 53, 55, 57, 58, 59, 61, 62, 65, 67, 69, 71, 73, 74, 77, 79, 82, 83, 85, 86, 87, 89, 91, 93, 94, 95, 97, 101, 103, 106, 107, 109, 111, 113, 115, 118

Offset: 1

N. J. A. Sloane

nonn

A001222(a(n)) <= 2; A054576(a(n)) = 1. - Reinhard Zumkeller, Mar 10 2006
Products of two noncomposite numbers. - Juri-Stepan Gerasimov, Apr 15 2010
Also, numbers with permutations of all divisors only with coprime adjacent elements: A109810(a(n)) > 0. - Reinhard Zumkeller, May 24 2010
A060278(a(n)) = 0. - Reinhard Zumkeller, Apr 05 2013
1 together with numbers k such that sigma(k) + phi(k) - d(k) = 2k - 2. - Wesley Ivan Hurt, May 03 2015
Products of two not necessarily distinct terms of A008578 (the same relation between A000040 and A001358). - Flávio V. Fernandes, May 28 2021

Felix Fröhlich, Table of n, a(n) for n = 1..10000 (first 1000 terms from Reinhard Zumkeller)
Andreas Weingartner, Uniform distribution of alpha*n modulo one for a family of integer sequences, arXiv:2303.16819 [math.NT], 2023.

Union of A008578 and A001358. Complement of A033942.
Cf. A063928, A001222, A054576, A109810, A060278.
A101040(a(n))=1 for n>1.
Subsequence of A037144. - Reinhard Zumkeller, May 24 2010
A098962 and A139690 are subsequences.

a037143 n = a037143_list !! (n-1)
a037143_list = 1 : merge a000040_list a001358_list where
   merge xs'@(x:xs) ys'@(y:ys) =
         if x < y then x : merge xs ys' else y : merge xs' ys
-- Reinhard Zumkeller, Dec 18 2012

with(numtheory): A037143:=n->`if`(bigomega(n)<3,n,NULL): seq(A037143(n), n=1..200); # Wesley Ivan Hurt, May 03 2015

Select[Range[120], PrimeOmega[#] <= 2 &] (* Ivan Neretin, Aug 16 2015 *)

is(n)=bigomega(n)<3 \\ Charles R Greathouse IV, Apr 29 2015

from math import isqrt
from sympy import primepi, primerange
def A037143(n):
    def f(x): return int(n-2+x-primepi(x)-sum(primepi(x//k)-a for a,k in enumerate(primerange(isqrt(x)+1))))
    kmin, kmax = 1,2
    while f(kmax) >= kmax:
        kmax <<= 1
    while True:
        kmid = kmax+kmin>>1
        if f(kmid) < kmid:
            kmax = kmid
        else:
            kmin = kmid
        if kmax-kmin <= 1:
            break
    return kmax # Chai Wah Wu, Aug 23 2024

More terms from Henry Bottomley, Aug 15 2001

A016038 Strictly non-palindromic numbers: n is not palindromic in any base b with 2 <= b <= n-2.

Original entry on oeis.org

0, 1, 2, 3, 4, 6, 11, 19, 47, 53, 79, 103, 137, 139, 149, 163, 167, 179, 223, 263, 269, 283, 293, 311, 317, 347, 359, 367, 389, 439, 491, 563, 569, 593, 607, 659, 739, 827, 853, 877, 977, 983, 997, 1019, 1049, 1061, 1187, 1213, 1237, 1367, 1433, 1439, 1447, 1459

Offset: 1

Robert G. Wilson v

All elements of the sequence greater than 6 are prime (ab = a(b-1) + a or a^2 = (a-1)^2 + 2(a-1) + 1). Mersenne and Fermat primes are not in the sequence.
Additional comments: if you can factor a number as a*b then it is a palindrome in base b-1, where b is the larger of the two factors. (If the number is a square, then it can be a palindrome in an additional way, in base (sqrt(n)-1)). The a*b form does not work when a = b-1, but of course there are no two consecutive primes (other than 2,3, which explains the early special cases), so if you can factor a number as a*(a-1), then another factorization also exists. - Michael B Greenwald (mbgreen(AT)central.cis.upenn.edu), Jan 01 2002
Note that no prime p is palindromic in base b for the range sqrt(p) < b < p-1. Hence to find non-palindromic primes, we need only examine bases up to floor(sqrt(p)), which greatly reduces the computational effort required. - T. D. Noe, Mar 01 2008
No number n is palindromic in any base b with n/2 <= b <= n-2, so this is also numbers not palindromic in any base b with 2 <= b <= n/2.
Sequence A047811 (this sequence without 0, 1, 2, 3) is mentioned in the Guy paper, in which he reports on unsolved problems. This problem came from Mario Borelli and Cecil B. Mast. The paper poses two questions about these numbers: (1) Can palindromic or nonpalindromic primes be otherwise characterized? and (2) What is the cardinality, or the density, of the set of palindromic primes? Of the set of nonpalindromic primes? - T. D. Noe, Apr 18 2011
From Robert G. Wilson v, Oct 22 2014 and Nov 03 2014: (Start)
Define f(n) to be the number of palindromic representations of n in bases b with 1 < b < n, see A135551.
For A016038, f(n) = 1 for all n. Only the numbers n = 0, 1, 4 and 6 are not primes.
For f(n) = 2, all terms are prime or semiprimes (prime omega <= 2 (A037143)) with the exception of 8 and 12;
For f(n) = 3, all terms are at most 3-almost primes (prime omega <= 3 (A037144)), with the exception of 16, 32, 81 and 625;
For f(n) = 4, all terms are at most 4-almost primes, with the exception of 64 and 243;
For f(n) = 5, all terms are at most 5-almost primes, with the exception of 128, 256 and 729;
For f(n) = 6, all terms are at most 6-almost primes, with the sole exception of 2187;
For f(n) = 7, all terms are at most 7-almost primes, with the exception of 512, 2048 and 19683; etc. (End)

Paul Guinand, Strictly non-palindromic numbers, unpublished note, 1996.

T. D. Noe, Table of n, a(n) for n = 1..10001
K. S. Brown, On General Palindromic Numbers
Patrick De Geest, Palindromic numbers beyond base 10
R. K. Guy, Conway's RATS and other reversals, Amer. Math. Monthly, 96 (1989), 425-428.
John P. Linderman, Description of A135549-A016038
John P. Linderman, Perl program [Use the command: HASNOPALINS=1 palin.pl]

Cf. A047811, A050812, A050813, A037183, A135550, A135551, A135549, A138348.

PalindromicQ[n_, base_] := FromDigits[Reverse[IntegerDigits[n, base]], base] == n; PalindromicBases[n_] := Select[Range[2, n-2], PalindromicQ[n, # ] &]; StrictlyPalindromicQ[n_] := PalindromicBases[n] == {}; Select[Range[150], StrictlyPalindromicQ] (* Herman Beeksma, Jul 16 2005*)
palindromicBases[n_] := Module[{p}, Table[p = IntegerDigits[n, b]; If[ p == Reverse[p], {b, p}, Sequence @@ {}], {b, 2, n - 2}]]; lst = {0, 1, 4, 6}; Do[ If[ Length@ palindromicBases@ Prime@n == 0, AppendTo[lst, Prime@n]], {n, 10000}]; lst (* Robert G. Wilson v, Mar 08 2008 *)
Select[Range@ 1500, Function[n, NoneTrue[Range[2, n - 2], PalindromeQ@ IntegerDigits[n, #] &]]] (* Michael De Vlieger, Dec 24 2017 *)

is(n)=!for(b=2,n\2,Vecrev(d=digits(n,b))==d&&return) \\ M. F. Hasler, Sep 08 2015

from itertools import count, islice
from sympy.ntheory.factor_ import digits
def A016038_gen(startvalue=0): # generator of terms >= startvalue
    return filter(lambda n: all((s := digits(n,b)[1:])[:(t:=len(s)+1>>1)]!=s[:-t-1:-1] for b in range(2,n-1)), count(max(startvalue,0)))
A016038_list = list(islice(A016038_gen(),30)) # Chai Wah Wu, Jan 17 2024

a(n) = A047811(n-4) for n > 4. - M. F. Hasler, Sep 08 2015

Extended and corrected by Patrick De Geest, Oct 15 1999

Edited by N. J. A. Sloane, Apr 09 2008

cp's OEIS Frontend

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