A367359 Value of A121805(k) for k = A367358(n).
1, 12, 35, 94, 135, 248, 331, 461, 530, 651, 744, 809, 908, 1068, 2070, 3039, 4093, 5012, 6013, 7042, 8026, 9055, 10058, 20047, 30092, 40017, 50008, 60054, 70063, 80065, 90022, 100058, 200051, 300060, 400093, 500028, 600074, 700013, 800022, 900055, 1000006, 2000047, 3000008, 4000061, 5000034, 6000055, 7000008, 8000056, 9000013, 10000036, 20000038, 30000029, 40000039, 50000030, 60000051, 70000064, 80000064, 90000077
Offset: 1
Programs
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Python
def agen(): # generator of terms n, an, y, prev_lead, prev_length = 1, 1, 1, None, None while y < 10: san = str(an) lead, length = san[0], len(san) if lead != prev_lead or length != prev_length: yield an an, y = an + 10*(an%10), 1 while y < 10: if str(an+y)[0] == str(y): an += y break y += 1 n, prev_lead, prev_length = n+1, lead, length print(list(agen())) # Michael S. Branicky, Nov 22 2023
Extensions
More than the usual number of terms are shown in order to match the data in A367358.
More terms from Michael S. Branicky, Nov 22 2023
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