A360640
a(n) is the start of the least run of exactly n consecutive odd numbers that are A000120-perfect numbers (A175522).
Original entry on oeis.org
25, 123, 31803, 8019811, 130194395
Offset: 1
Table of values of A000120 and A093653 for k = a(n), a(n)+2, ..., a(n)+2*(n-1):
n | a(n) A000120(k) A093653(k)
--+----------------------------------------------------------
1 | 25 3 6
2 | 123 6, 6. 12, 12
3 | 31803 10, 10, 11 20, 20, 22
4 | 8019811 15, 15, 16, 15 30, 30, 32, 30
5 | 130194395 17, 17, 18, 15, 16, 16 34, 34, 36, 30, 32, 32
-
q[n_] := DivisorSum[n, DigitCount[#, 2, 1] &] == 2 * DigitCount[n, 2, 1]; seq[len_, nmax_] := Module[{s = Table[0, {len}], v = {}, n = 1, c = 0, m}, While[c <= len && n <= nmax, If[q[n], v = Join[v, {n}], m = Length[v]; v = {}; If[0 < m <= len && s[[m]] == 0, c++; s[[m]] = n - 2*m]]; n += 2]; s]; seq[3, 10^5]
-
lista(len, nmax) = {my(s = vector(len), v=[], n = 1, c = 0, m); while(c <= len && n <= nmax, if(sumdiv(n, d, hammingweight(d)) == 2 * hammingweight(n), v = concat(v, n), m =#v; v = []; if(0 < m && m <= len && s[m] == 0, c++; s[m] = n - 2*m)); n += 2); s};
A373094
a(n) is the least number k such that A373092(k) = n.
Original entry on oeis.org
1, 4, 7, 12, 24, 120, 1260, 1829520
Offset: 0
The iterations for the n = 0..7 are:
n a(n) iterations
- ------- --------------------------------------------------
0 1 1
1 4 4 -> 3
2 7 7 -> 4 -> 3
3 12 12 -> 9 -> 5 ->3
4 24 24 -> 12 -> 9 -> 5 -> 3
5 120 120 -> 36 -> 15 -> 9 -> 5 -> 3
6 1260 1260 -> 120 -> 36 -> 15 -> 9 -> 5 -> 3
7 1829520 1829520 -> 1260 -> 120 -> 36 -> 15 -> 9 -> 5 -> 3
-
d[n_] := d[n] = DivisorSum[n, Plus @@ IntegerDigits[#, 2] &];
f[n_] := -2 + Length@ FixedPointList[d, n];
seq[len_] := Module[{s = Table[0, {len}], c = 0, i, n = 1}, While[c < len, i = f[n] + 1; If[i <= len && s[[i]] == 0, c++; s[[i]] = n]; n++]; s]; seq[7]
-
f(n) = {my(c = 0); while(6 % n, n = sumdiv(n, d, hammingweight(d)); c++); c;}
lista(len) = {my(s = vector(len), c = 0, i, n = 1); while(c < len, i = f(n) + 1; if(i <= len && s[i] == 0, c++; s[i] = n); n++); s;}
A339551
Starts of runs of 3 consecutive numbers with the same product of the binary weights of their divisors (A339549).
Original entry on oeis.org
513059433, 3007912105, 4791685641, 11555664153, 44615854297, 111890605585, 121111724905, 163901238153
Offset: 1
513059433 is a term since A339549(513059433) = A339549(513059434) = A339549(513059435) = 1166400.
-
f[n_] := Times @@ (DigitCount[#, 2, 1] & /@ Divisors[n]); s = {}; m = 3; fs = f /@ Range[m]; Do[If[Equal @@ fs, AppendTo[s, n - m]]; fs = Rest @ AppendTo[fs, f[n]], {n, m + 1, 5*10^9}]; s
A360643
a(n) is the least A000120-perfect number (A175522) whose binary weight (A000120) is n, or 0 if no such number exists.
Original entry on oeis.org
2, 0, 25, 169, 841, 95, 247, 943, 767, 5999, 6139, 16123, 30655, 90109, 122847, 245695, 522237, 1572591, 1966015, 3932095, 12582651, 28311519, 33423343, 100663023, 133693435, 402128831, 931135479, 1069547515, 1610612607, 11802771447, 12884901567, 25736249279
Offset: 1
a(1) = 2 since A000120(2) = 1 and A093653(2)/A000120(2) = 4/2 = 2.
a(2) = 0 since there is no number m with binary weight 2 and with A093653(m) = 4.
a(3) = 25 since A000120(25) = 3 and A093653(25)/A000120(25) = 6/3 = 2, and 25 is the least number with this property.
-
seq[len_, nmax_] := Module[{s = Table[-1, {len}], n = 3, c = 2, bw, dbw}, s[[1]] = 2; While[c < len && n <= nmax, bw = DigitCount[n, 2, 1]; If[bw <= len && s[[bw]] < 0, dbw = DivisorSum[n, DigitCount[#, 2, 1] &]; If[dbw == 2*bw, c++; s[[bw]] = n]]; n += 2]; s]; seq[16, 10^6]
-
lista(len, nmax) = {my(s = vector(len,i,-1), n = 3, c = 2, bw, dbw); s[1] = 2; while(c < len && n <= nmax, bw = hammingweight(n); if(bw <= len && s[bw] < 0, dbw = sumdiv(n, d, hammingweight(d)); if(dbw == 2*bw, c++; s[bw] = n)); n += 2); s};
Comments