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(12) = 1*2 + 2*1 = 4, since 12 = 2^2 *3^1 = (p_1)^2 *(p_2)^1.
a056239 n = sum $ zipWith (*) (map a049084 $ a027748_row n) (a124010_row n) -- Reinhard Zumkeller, Apr 27 2013
# To get 10000 terms. First make prime table: M:=10000; pl:=array(1..M); for i from 1 to M do pl[i]:=0; od: for i from 1 to M do if ithprime(i) > M then break; fi; pl[ithprime(i)]:=i; od: # Decode Maple's amazing syntax for factoring integers: g:=proc(n) local e,p,t1,t2,t3,i,j,k; global pl; t1:=ifactor(n); t2:=nops(t1); if t2 = 2 and whattype(t1) <> `*` then p:=op(1,op(1,t1)); e:=op(2,t1); t3:=pl[p]*e; else t3:=0; for i from 1 to t2 do j:=op(i,t1); if nops(j) = 1 then e:=1; p:=op(1,j); else e:=op(2,j); p:=op(1,op(1,j)); fi; t3:=t3+pl[p]*e; od: fi; t3; end; # N. J. A. Sloane, May 10 2006 A056239 := proc(n) add( numtheory[pi](op(1,p))*op(2,p), p = ifactors(n)[2]) ; end proc: # R. J. Mathar, Apr 20 2010 # alternative: with(numtheory): a := proc (n) local B: B := proc (n) local nn, j, m: nn := op(2, ifactors(n)): for j to nops(nn) do m[j] := op(j, nn) end do: [seq(seq(pi(op(1, m[i])), q = 1 .. op(2, m[i])), i = 1 .. nops(nn))] end proc: add(B(n)[i], i = 1 .. nops(B(n))) end proc: seq(a(n), n = 1 .. 130); # Emeric Deutsch, May 19 2015
a[1] = 0; a[2] = 1; a[p_?PrimeQ] := a[p] = PrimePi[p]; a[n_] := a[n] = Total[#[[2]]*a[#[[1]]] & /@ FactorInteger[n]]; a /@ Range[91] (* Jean-François Alcover, May 19 2011 *) Table[Total[FactorInteger[n] /. {p_, c_} /; p > 0 :> PrimePi[p] c], {n, 91}] (* Michael De Vlieger, Jul 12 2017 *)
A056239(n) = if(1==n,0,my(f=factor(n)); sum(i=1, #f~, f[i,2] * primepi(f[i,1]))); \\ Antti Karttunen, Oct 26 2014, edited Jan 13 2020
from sympy import primepi, factorint def A056239(n): return sum(primepi(p)*e for p, e in factorint(n).items()) # Chai Wah Wu, Jan 01 2023
(require 'factor) ;; Uses the function factor available in Aubrey Jaffer's SLIB Scheme library. (define (A056239 n) (apply + (map A049084 (factor n)))) ;; Antti Karttunen, Oct 26 2014
The subset {2,3,6} has 7 = 2*2 + 1*3 + 0*6 so is counted under a(7).
The a(1) = 1 through a(4) = 14 subsets:
{1} {1} {1} {1}
{2} {3} {2}
{1,2} {1,2} {4}
{1,3} {1,2}
{2,3} {1,3}
{1,2,3} {1,4}
{2,3}
{2,4}
{3,4}
{1,2,3}
{1,2,4}
{1,3,4}
{2,3,4}
{1,2,3,4}
combs[n_,y_]:=With[{s=Table[{k,i},{k,y},{i,0,Floor[n/k]}]},Select[Tuples[s],Total[Times@@@#]==n&]];
Table[Length[Select[Subsets[Range[n]],combs[n,#]!={}&]],{n,0,5}]
a(n)={
my(comb(k,b)=while(b>>k, b=bitor(b, b>>k); k*=2); b);
my(recurse(k,b)=
if(bittest(b,0), 2^(n+1-k),
if(2*k>n, 2^(n+1-k) - 2^sum(j=k, n, !bittest(b,j)),
self()(k+1, b) + self()(k+1, comb(k,b)) )));
recurse(1, 1<Andrew Howroyd, Sep 04 2023
The set {4,5,6} cannot be linearly combined to obtain 7 so is counted under a(7), but we have 8 = 2*4 + 0*5 + 0*6, so it is not counted under a(8).
The a(1) = 1 through a(8) = 12 subsets:
{} {} {} {} {} {} {} {}
{2} {3} {2} {4} {2} {3}
{3} {5} {3} {5}
{4} {4,5} {4} {6}
{2,4} {5} {7}
{3,4} {6} {3,6}
{2,4} {3,7}
{2,6} {5,6}
{3,5} {5,7}
{3,6} {6,7}
{4,5} {3,6,7}
{4,6} {5,6,7}
{5,6}
{2,4,6}
{3,5,6}
{4,5,6}
combs[n_,y_]:=With[{s=Table[{k,i},{k,y},{i,0,Floor[n/k]}]},Select[Tuples[s],Total[Times@@@#]==n&]];
Table[Length[Select[Subsets[Range[n-1]],combs[n,#]=={}&]],{n,5}]
The partition (4,2,1) has 3 = (2)+(1) or 3 = (1+1+1) so is counted under a(7).
The a(1) = 1 through a(7) = 13 partitions:
(1) (11) (21) (22) (32) (42) (52)
(111) (31) (41) (51) (61)
(211) (221) (321) (322)
(1111) (311) (411) (331)
(2111) (2211) (421)
(11111) (3111) (511)
(21111) (2221)
(111111) (3211)
(4111)
(22111)
(31111)
(211111)
(1111111)
combs[n_,y_]:=With[{s=Table[{k,i},{k,y}, {i,0,Floor[n/k]}]}, Select[Tuples[s], Total[Times@@@#]==n&]];
Table[Length[Select[IntegerPartitions[n], combs[Length[#], Union[#]]!={}&]], {n,0,15}]
3 cannot be written as a nonnegative linear combination of 2 and 5, so (5,2,2) is counted under a(9).
The a(2) = 1 through a(10) = 7 partitions:
(2) (3) (4) (5) (6) (7) (8) (9) (10)
(3,3) (4,3) (4,4) (5,4) (5,5)
(2,2,2) (5,3) (6,3) (6,4)
(4,2,2) (5,2,2) (7,3)
(4,4,2)
(6,2,2)
(2,2,2,2,2)
combs[n_,y_]:=With[{s=Table[{k,i},{k,y},{i,0,Floor[n/k]}]},Select[Tuples[s],Total[Times@@@#]==n&]];
Table[Length[Select[IntegerPartitions[n],combs[Length[#],Union[#]]=={}&]],{n,0,15}]
The a(3) = 1 through a(10) = 7 strict partitions:
(2,1) (3,1) (3,2) (4,2) (5,2) (6,2) (7,2) (8,2)
(4,1) (5,1) (6,1) (7,1) (8,1) (9,1)
(3,2,1) (4,2,1) (4,3,1) (4,3,2) (5,3,2)
(5,2,1) (5,3,1) (5,4,1)
(6,2,1) (6,3,1)
(7,2,1)
(4,3,2,1)
combs[n_,y_]:=With[{s=Table[{k,i},{k,y}, {i,0,Floor[n/k]}]}, Select[Tuples[s], Total[Times@@@#]==n&]];
Table[Length[Select[IntegerPartitions[n], UnsameQ@@#&&combs[Length[#], Union[#]]!={}&]], {n,0,15}]
The a(2) = 1 through a(16) = 10 strict partitions (A..G = 10..16):
2 3 4 5 6 7 8 9 A B C D E F G
43 53 54 64 65 75 76 86 87 97
63 73 74 84 85 95 96 A6
83 93 94 A4 A5 B5
542 642 A3 B3 B4 C4
652 752 C3 D3
742 842 654 754
762 862
852 952
942 A42
combs[n_,y_]:=With[{s=Table[{k,i},{k,y}, {i,0,Floor[n/k]}]}, Select[Tuples[s], Total[Times@@@#]==n&]];
Table[Length[Select[IntegerPartitions[n], UnsameQ@@#&&combs[Length[#], Union[#]]=={}&]], {n,0,30}]
The set {1,2,4} has 3 = (2)+(1) or 3 = (1+1+1) so is counted under a(4).
The a(0) = 1 through a(4) = 12 subsets:
{} {} {} {} {}
{1} {1} {1} {1}
{1,2} {1,2} {1,2}
{1,3} {1,3}
{2,3} {1,4}
{1,2,3} {2,3}
{2,4}
{1,2,3}
{1,2,4}
{1,3,4}
{2,3,4}
{1,2,3,4}
combs[n_,y_]:=With[{s=Table[{k,i},{k,y}, {i,0,Floor[n/k]}]}, Select[Tuples[s], Total[Times@@@#]==n&]];
Table[Length[Select[Subsets[Range[n]], combs[Length[#], Union[#]]!={}&]], {n,0,10}]
from itertools import combinations from sympy.utilities.iterables import partitions def A367222(n): c, mlist = 1, [] for m in range(1,n+1): t = set() for p in partitions(m): t.add(tuple(sorted(p.keys()))) mlist.append([set(d) for d in t]) for k in range(1,n+1): for w in combinations(range(1,n+1),k): ws = set(w) for s in mlist[k-1]: if s <= ws: c += 1 break return c # Chai Wah Wu, Nov 16 2023
3 cannot be written as a nonnegative linear combination of 2, 4, and 5, so {2,4,5} is counted under a(6).
The a(2) = 1 through a(6) = 15 subsets:
{2} {2} {2} {2} {2}
{3} {3} {3} {3}
{4} {4} {4}
{3,4} {5} {5}
{3,4} {6}
{3,5} {3,4}
{4,5} {3,5}
{2,4,5} {3,6}
{4,5}
{4,6}
{5,6}
{2,4,5}
{2,4,6}
{2,5,6}
{4,5,6}
combs[n_,y_]:=With[{s=Table[{k,i},{k,y}, {i,0,Floor[n/k]}]}, Select[Tuples[s], Total[Times@@@#]==n&]];
Table[Length[Select[Subsets[Range[n]], combs[Length[#],Union[#]]=={}&]], {n,0,10}]
from itertools import combinations from sympy.utilities.iterables import partitions def A367223(n): c, mlist = 0, [] for m in range(1,n+1): t = set() for p in partitions(m): t.add(tuple(sorted(p.keys()))) mlist.append([set(d) for d in t]) for k in range(1,n+1): for w in combinations(range(1,n+1),k): ws = set(w) for s in mlist[k-1]: if s <= ws: break else: c += 1 return c # Chai Wah Wu, Nov 16 2023
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