This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
a[n_] := a[n] = Block[{k = 2, b = a[n - 1], c = 2 Range[0, 4]}, While[Mod[k*b,10]==0 || Union@ Join[c, IntegerDigits[k*b]] != c, k++]; k*b]; a[1] = 2; Array[a,14] (* Robert G. Wilson v, May 26 2014 *)
Join[{2,4,8},Flatten[Table[{24,48}10^n,{n,0,20}]]] (* Harvey P. Dale, May 01 2013 *)
a(9) = A078223(10) / A078223(9) = 26266464 / 260064 = 101.
a(6) = 567 = 7*a(5); the digits alternate odd, even, odd.
isA030141 := proc(n) local dgs,i ; dgs := convert(n,base,10) ; for i from 1 to nops(dgs)-1 do if ( op(i,dgs)+op(i+1,dgs)) mod 2 = 0 then RETURN(false) ; fi ; od ; RETURN(true) ; end: A078226 := proc(nmax) local a,f; a := [1] ; while nops(a) < nmax do f := 3 ; while true do if isA030141(f*op(-1,a)) then a := [op(a),f*op(-1,a)] ; print(op(-1,a)) ; break ; fi ; f := f+2 ; od ; od ; end: A078226(13) ; # R. J. Mathar, Mar 01 2007
A078226_list = [1] for _ in range(20): x = A078226_list[-1] y, x2 = x, 2*x while True: y += x2 s = str(y) for j in range(len(s)-1, -1, -2): if not s[j] in ('1', '3', '5', '7', '9'): break else: for k in range(len(s)-2, -1, -2): if not s[k] in ('0', '2', '4', '6', '8'): break else: A078226_list.append(y) break # Chai Wah Wu, Nov 06 2014
a(7) = 672 = 7*a(6) = 7*96. Starting with the unit digit the digits in 672 are alternately even and odd.
isAltr := proc(n) local nshft,osgn,sgn ; nshft := n ; osgn := ( n mod 10 ) mod 2 ; while nshft >= 10 do nshft := floor(nshft/10) ; sgn := ( nshft mod 10 ) mod 2 ; if sgn = osgn then RETURN(false) ; fi ; osgn := sgn ; od ; RETURN(true) ; end: A078227 := proc(prev) local m; m := 2 ; while true do if isAltr(m*prev) then RETURN(m*prev) ; fi ; m := m+1 ; od ; end: n := 2 : while true do print(n) ; n := A078227(n) : od : # R. J. Mathar, Nov 12 2006
A078227_list = [2] for _ in range(20): x = A078227_list[-1] y = x while True: y += x s = str(y) for j in range(len(s)-1,-1,-2): if not s[j] in ('0','2','4','6','8'): break else: for k in range(len(s)-2,-1,-2): if not s[k] in ('1','3','5','7','9'): break else: A078227_list.append(y) break # Chai Wah Wu, Nov 06 2014
a(10) = 315 = 21 * 15 is the smallest multiple of 21 which contains only odd digits. a(4998) = 33339995 = 9997 * 3335 is the smallest multiple of 9997 which contains only odd digits, so this is the answer to the IMTS problem.
a[n_] := Module[{m = 2*n + 1, k}, k = 3*m; While[!AllTrue[IntegerDigits[k], OddQ], k += 2*m]; k]; Array[a, 50, 0] (* Amiram Eldar, Jan 04 2022 *)
isok(k) = my(d=digits(k)); #d == #select(x->((x%2)==1), d); a(n) = my(k=6*n+3); while (!isok(k), k+=4*n+2); k; \\ Michel Marcus, Jan 04 2022
from itertools import product, count def A350536(n): m = 2*n+1 for l in count(len(str(m))): for s in product('13579',repeat=l): k = int(''.join(s)) if k > m and k % m == 0: return k # Chai Wah Wu, Jan 11 2022
The smallest proper multiple of 21 = 2*10+1 with only odd digits is A350536(10) = 315, as 315 = 21 * 15, a(10) = 15.
Table[m=2;While[Or@@EvenQ[IntegerDigits[(2n+1)*++m]]];m,{n,0,79}] (* Giorgos Kalogeropoulos, Jan 12 2022 *)
isok(k) = my(d=digits(k)); #d == #select(x->((x%2)==1), d); a(n) = my(k=6*n+3); while (!isok(k), k+=4*n+2); k/(2*n+1); \\ Michel Marcus, Jan 12 2022
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