A344488 Numbers that start a product crescendo of record length.
1, 2, 3, 7, 47, 181, 1307, 2503, 40973, 46833, 109177, 2885373, 11744311, 192968969, 899988745
Offset: 1
Examples
181 is in the list because it begins a product crescendo that is longer than any beginning at any smaller number. Here is the crescendo: 1 * 181 = 181 2 * 91 = 182 3 * 61 = 183 4 * 46 = 184 5 * 37 = 185 6 * 31 = 186 11 * 17 = 187 47 * 4 = 188 63 * 3 = 189 95 * 2 = 190 191 * 1 = 191 This set of 11 products forms a longer crescendo than the previous record (which started at 47), and is the longest until the set of 13 products it is possible to write starting from 1307 (the next entry in the sequence). Additional example: the crescendo from 2885373 (length 27) goes: 1 * 2885373 = 2885373 2 * 1442687 = 2885374 5 * 577075 = 2885375 6 * 480896 = 2885376 11 * 262307 = 2885377 19 * 151862 = 2885378 21 * 137399 = 2885379 89 * 32420 = 2885380 859 * 3359 = 2885381 1458 * 1979 = 2885382 4817 * 599 = 2885383 12437 * 232 = 2885384 19365 * 149 = 2885385 33551 * 86 = 2885386 93077 * 31 = 2885387 131154 * 22 = 2885388 221953 * 13 = 2885389 288539 * 10 = 2885390 320599 * 9 = 2885391 360674 * 8 = 2885392 412199 * 7 = 2885393 480899 * 6 = 2885394 577079 * 5 = 2885395 721349 * 4 = 2885396 961799 * 3 = 2885397 1442699 * 2 = 2885398 2885399 * 1 = 2885399
Crossrefs
Cf. A038507.
Programs
-
PARI
b(n)={if(n==1, 1, my(m=1); for(k=1, oo, fordiv(n+k, d, if(d>m, m=d; break)); if(m==n+k, return(k+1))))} lista(lim)={my(m=0); for(n=1, lim, my(t=b(n)); if(t > m, print1(n, ", "); m=t))} \\ Andrew Howroyd, May 21 2021
Extensions
a(13)-a(14) from Rémy Sigrist, May 21 2021
a(15) from Jon E. Schoenfield, May 21 2021
Comments